[OI - 高精度]高精度模板

1.高精度加法

復雜度O(n)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
#define DEBUG(x) cerr << #x << "=" << x << endl
const int L = 110; 

string a, b;

string add(string a, string b)//只限兩個非負整數相加 
{
    string ans;
    int na[L] = {0};
    int nb[L] = {0};
    int la = a.size();
    int lb = b.size();
    for (int i = 0; i < la; i++) 
        na[la - 1 - i] = a[i] - '0';
    for (int i = 0; i < lb; i++) 
        nb[lb - 1 - i] = b[i] - '0';
    int lmax = la > lb ? la : lb;
    for (int i = 0; i < lmax; i++) 
    {
        na[i] += nb[i];
        na[i + 1] += na[i] / 10;
        na[i] %= 10;
    } 
    if (na[lmax]) lmax++;
    for (int i = lmax - 1; i >= 0; i--) 
        ans += na[i] + '0';
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> a >> b) 
        cout << add(a, b) << endl;
    return 0;
}

2.高精度減法

復雜度O(n)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring> 

using namespace std;
//#define DEBUG(x) cerr << #x << "=" << x << endl
typedef unsigned char BYTE;

BYTE a[10005] = {0};
BYTE b[10005] = {0};
BYTE *pa = 0, *pb = 0;
int la, lb;
int sign;

int main()
{
    scanf("%s", a);
    scanf("%s", b);
    la = strlen((char*)a);
    lb = strlen((char*)b);
    int i;
    for (int i = 0; i < la; i++)
    {
        a[i] -= '0';
    }
    for (int i = 0; i < lb; i++)
    {
        b[i] -= '0';
    }
    if (la > lb)
    {
        sign = 1;
    }
    else if (la < lb)
    {
        sign = 0;
    }
    else 
    {
        sign = -1;
        int BCT;
        for (BCT = 0; BCT < la; BCT++)
        {
            if (a[BCT] > b[BCT])
            {
                sign = 1;
                break;
            }
            if (a[BCT] < b[BCT])
            {
                sign = 0;
                break;
            }
        }
        if (sign == -1)
        {
            printf("0");
            return 0;
        }
    }
    if (sign == 1)
    {
        pa = a;
        pb = b;
    }
    else 
    {
        pa = b;
        pb = a;
        int buf;
        buf = la;
        la = lb;
        lb = buf;
    }
    memmove(pb + la - lb, pb, lb);
    memset(pb, 0, la - lb);
    for (int i = 0; i < la; i++)
    {
        pb[i] = 9 - pb[i];
    }
    pb[la - 1]++;
    for (int i = la - 1; i >= 0; i--)
    {
        pa[i] += pb[i];
        if (pa[i] >= 10)
        {
            pa[i] -= 10;
            if (i != 0)
            {
                pa[i - 1]++;
            }
        }
    }
    for (int i = 0; i < la; i++)
    {
        pa[i] += '0';
    }
    if (sign == 0)
    {
        printf("-");
    }
    for (int i = 0; i < la; i++)
    {
        if (pa[i] != '0')
        {
            printf((char*)pa + i);
            return 0;
        }
    }
    return 0;
}

3.高精度乘法

復雜度O(n*n)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
//#define DEBUG(x) cerr << #x << "=" << x << endl
const int L = 110;

string a, b;

string mul(string a, string b)
{
    string s;
    int na[L];
    int nb[L];
    int nc[L];
    int La = a.size();
    int Lb = b.size();
    fill(na, na + L, 0);
    fill(nb, nb + L, 0);
    fill(nc, nc + L, 0);
    for (int i = La - 1; i >= 0; i--)
        na[La - i] = a[i] - '0';
    for (int i = Lb - 1; i >= 0; i--)
        nb[Lb - i] = b[i] - '0';
    for (int i = 1; i <= La; i++)
    {
        for (int j = 1; j <= Lb; j++)
        {
            nc[i + j - 1] += na[i] * nb[j];
        }
    }
    for (int i = 1; i <= La + Lb; i++)
    {
        nc[i + 1] += nc[i] / 10;
        nc[i] %= 10;
    }
    if (nc[La + Lb]) 
       s += nc[La + Lb] + '0';
    for (int i = La + Lb - 1; i >= 1; i--)
        s += nc[i] + '0';
    return s;  
}

int main()
{
    while (cin >> a >> b) 
        cout << mul(a, b) << endl;
    return 0;
}

4.高精度乘法FFT優化

復雜度O(nlogn)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
//#define DEBUG(x) cerr << #x << "=" << x << endl
#define L(x) (1 << (x))
const double PI = acos(-1.0);
const int maxn = 133015;

double ax[maxn], ay[maxn], bx[maxn], by[maxn];
char sa[maxn / 2], sb[maxn / 2];
int sum[maxn];
int x1[maxn], x2[maxn];
string a, b;

int max(int a, int b)
{
    return a > b ? a : b;
}

int revv(int x, int bits)
{
    int ret = 0;
    for (int i = 0; i < bits; i++)
    {
        ret <<= 1;
        ret |= x & 1;
        x >>= 1;
    }
    return ret;
}

void fft(double *a, double *b, int n, bool rev)
{
    int bits = 0;
    while (1 << bits < n) 
        ++bits;
    for (int i = 0; i < n; i++)
    {
        int j = revv(i, bits);
        if (i < j)
        {
            swap(a[i], a[j]);
            swap(b[i], b[j]);
        }
    }
    for (int len = 2; len <= n; len <<= 1)
    {
        int half = len >> 1;
        double wmx = cos(2 * PI / len);
        double wmy = sin(2 * PI / len);
        if (rev)
            wmy = -wmy;
        for (int i = 0; i < n; i += len)
        {
            double wx = 1;
            double wy = 0;
            for (int j = 0; j < half; j++)
            {
                double cx = a[i + j];
                double cy = b[i + j];
                double dx = a[i + j + half];
                double dy = b[i + j + half];
                double ex = dx * wx - dy * wy;
                double ey = dx * wy + dy * wx;
                a[i + j] = cx + ex;
                b[i + j] = cy + ey;
                a[i + j + half] = cx - ex;
                b[i + j + half] = cy - ey;
                double wnx = wx * wmx - wy * wmy;
                double wny = wx * wmy + wy * wmx;
                wx = wnx;
                wy = wny;
            }
        }
    }
    if (rev)
    {
        for (int i = 0; i < n; i++)
        {
            a[i] /= n;
            b[i] /= n;
        }
    }
}

int solve(int a[], int na, int b[], int nb, int ans[])
{
    int len = max(na, nb);
    int ln;
    for (int ln = 0; L(ln) < len; ++ln)
        len = L(++ln);
    for (int i = 0; i < len; ++i)
    {
        if (i >= na)
        {
            ax[i] = 0;                                                            
            ay[i] = 0;
        }             
        else
        {
            ax[i] = a[i];
            ay[i] = 0;
        }                         
    }
    fft(ax, ay, len, 0);
    for (int i = 0; i < len; ++i)
    {
        if (i >= nb)
        {
            bx[i] = 0;
            by[i] = 0;
        }
        else 
        {
            bx[i] = b[i];
            by[i] = 0;
        }
    }
    fft(bx, by, len, 0);
    for (int i = 0; i < len; ++i)
    {
        double cx = ax[i] * bx[i] - ay[i] * by[i];
        double cy = ax[i] * by[i] + ay[i] * bx[i];
        ax[i] = cx;
        ay[i] = cy;
    }
    fft(ax, ay, len, 1);
    for (int i = 0; i < len; ++i)
        ans[i] = (int)(ax[i] + 0.5);
    return len;
}

string mul(string sa, string sb)
{
    int l1, l2, l;
    int i;
    string ans;
    memset(sum, 0, sizeof(sum));
    l1 = sa.size();
    l2 = sb.size();
    for (int i = 0; i < l1; i++)
        x1[i] = sa[l1 - i - 1] - '0';
    for (int i = 0; i < l2; i++)
        x2[i] = sb[l2 - i - 1] - '0';
    l = solve(x1, l1, x2, l2, sum);
    for (int i = 0; i < l || sum[i] >= 10; i++)
    {
        sum[i + 1] += sum[i] / 10;
        sum[i] %= 10;
    }
    l = i;
    while (sum[l] <= 0 && l > 0)
        l--;
    for (int i = l; i >= 0; i--)
        ans += sum[i] + '0';
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> a >> b)
        cout << mul(a, b) << endl;
    return 0;
}

5.高精度乘單精度乘法

復雜度O(n)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
//#define DEBUG(x) cerr << #x << "=" << x << endl
const int L = 1e5 + 7;

int na[L];
string a;
int b;

string mul(string a, int b)
{
    string ans;
    int La = a.size();
    fill(na, na + L, 0);
    for (int i = La - 1; i >= 0; i--)
    na[La - i - 1] = a[i] - '0';
    int w = 0;
    for (int i = 0; i < La; i++)
    {
        na[i] = na[i] * b + w;
        w = na[i] / 10;
        na[i] = na[i] % 10;
    }
    while (w)
    {
        na[La++] = w % 10;
        w /= 10;
    }
    La--;
    while (La >= 0) 
        ans += na[La--] + '0';
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> a >> b) 
        cout << mul(a, b) << endl;
    return 0;
}

6.高精度除法(包括取模)

 復雜度O(n*n)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
//#define DEBUG(x) cerr << #x << "=" << x << endl
const int L = 110;

string a, b;

int sub(int *a, int *b, int La, int Lb)
{
    if (La < Lb) return -1;
    if (La == Lb)
    {
        for (int i = La - 1; i >= 0; i--)
        {
            if (a[i] > b[i]) break;
            else if (a[i] < b[i]) return -1;
        }
    }
    for (int i = 0; i < La; i++)
    {
        a[i] -= b[i];
        if (a[i] < 0) 
        {
            a[i] += 10;
            a[i + 1]--;
        }
    }
    for (int i = La - 1; i >= 0; i--)
        if (a[i]) return i + 1;
    return 0;
}

string div(string n1, string n2, int nn)
{
    string s, v;
    int a[L], b[L], r[L];
    int La = n1.size(), Lb = n2.size();
    int i, tp = La;
    fill(a, a + L, 0);
    fill(b, b + L, 0);
    fill(r, r + L, 0);
    for (int i = La - 1; i >= 0; i--) a[La - i - 1] = n1[i] - '0';
    for (int i = Lb - 1; i >= 0; i--) b[Lb - i - 1] = n2[i] - '0';
    if (La < Lb || (La == Lb && n1 < n2))
    {
        return n1;
    }
    int t = La - Lb;
    for (int i = La - 1;i >= 0; i--)
        if (i >= t) b[i] = b[i - t];
        else b[i] = 0;
    Lb = La;
    for (int j = 0; j <= t; j++)
    {
        int temp;
        while ((temp = sub(a, b + j, La, Lb - j)) >= 0)
        {
            La = temp;
            r[t - j]++;
        }
    }
    for (int i = 0; i < L - 10; i++)
    {
        r[i + 1] += r[i] / 10;
        r[i] %= 10;
    }
    while (!r[i]) i--;
    while (i >= 0) s += r[i--] + '0';
    i = tp;
    while (!a[i]) i--;
    while (i >= 0) v += a[i--] + '0';
    if (v.empty()) v = "0";
    if (nn == 1) return s;
    if (nn == 2) return v; 
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> a >> b) 
        cout << div(a, b, 1) << endl;
    return 0;
}

7.高精度除單精度除法

復雜度O(n)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
//#define DEBUG(x) cerr << #x << "=" << x << endl

string a;
int b;

string div(string a, int b)
{
    string r, ans;
    int d = 0;
    if (a == "0") return a;
    for (int i = 0; i < a.size(); i++)
    {
        r += (d * 10 + a[i] - '0') / b + '0';
        d = (d * 10  + (a[i] - '0')) % b;
    }
    int p = 0;
    for (int i = 0; i < r.size(); i++)
        if (r[i] != '0')
        {
            p = i;
            break;
        }
    return r.substr(p);
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> a >> b)
    {
        cout << div(a, b) << endl;
    }
    return 0;
}

8.高精度對單精度取模

復雜度O(n)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
//#define DEBUG(x) cerr << #x << "=" << x << endl

string a;
int b;

int mod(string a, int b)
{
    int d = 0;
    for (int i = 0; i < a.size(); i++) 
        d = (d * 10 + (a[i] - '0')) % b;
    return d;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> a >> b)
    {
        cout << mod(a, b) << endl;
    }
    return 0;
}

9.高精度階乘

 復雜度O(n*n)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
//#define DEBUG(x) cerr << #x << "=" << x << endl
const int L = 1e5 + 7;

int a[L];
int n;

string fac(int n)
{
    string ans;
    if (n == 0) return "1";
    fill(a, a + L, 0);
    int s = 0;
    int m = n;
    while (n)
    {
        a[++s] = m % 10;
        m /= 10;
    }
    for (int i = n - 1; i >= 2; i--)
    {
        int w = 0;
        for (int j = 1; j <= s; j++)
        {
            a[j] = a[j] * i + w;
            w = a[j] / 10;
            a[j] = a[j] % 10;
        }
    }
    while (!a[s]) s--;
    while (s >= 1) ans += a[s--] + '0';;
    return ans;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> n)
        cout << fac(n) << endl;
    return 0;
}

10.高精度冪

 復雜度O(nlognlogm)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;
//#define DEBUG(x) cerr << #x << "=" << x << endl
#define L(x) (1 << (x))
const double PI = acos(-1.0);
const int maxn = 122515;

double ax[maxn], ay[maxn], bx[maxn], by[maxn];
char sa[maxn / 2], sb[maxn / 2];
int sum[maxn];
int b;
int x1[maxn], x2[maxn];
string a;

int revv(int x, int bits)
{
    int ret = 0;
    for (int i = 0; i < bits; i++)
    {
        ret <<= 1;
        ret |= x & 1;
        x >>= 1;
    }
    return ret;
}

void fft(double *a, double *b, int n, bool rev)
{
    int bits = 0;
    while (1 << bits < n) ++bits;
    for (int i = 0; i < n; i++)
    {
        int j = revv(i, bits);
        if (i < j)
        {
            swap(a[i], a[j]);
            swap(b[i], b[j]);
        }
    }
    for (int len = 2; len <= n; len <<= 1)
    {
        int half = len >> 1;
        double wmx = cos(PI * 2 / len);
        double wmy = sin(PI * 2 / len);
        if (rev) wmy = -wmy;
        for (int i = 0; i < n; i += len)
        {
            double wx = 1;
            double wy = 0;
            for (int j = 0; j < half; j++)
            {
                double cx = a[i + j];
                double cy = b[i + j];
                double dx = a[i + j + half];
                double dy = a[i + j + half];
                double ex = dx * wx - dy * wy;
                double ey = dx * wy + dy * wx;
                a[i + j] = cx + ex;
                b[i + j] = cy + ey;
                a[i + j + half] = cx - ex;
                a[i + j + half] = cy - ey;
                double wnx = wx * wmx - wy * wmy;
                double wny = wx * wmy + wy * wmx;
                wx = wnx;
                wy = wny;
            }
        }
    }
    if (rev)
    {
        for (int i = 0; i < n; i++)
        {
            a[i] /= n;
            b[i] /= n; 
        }
    }
}

int solve(int a[], int na, int b[], int nb, int ans[])  
{  
    int len = max(na, nb);
    int ln;  
    for (ln = 0; L(ln) < len; ++ln);  
        len = L(++ln);  
    for (int i = 0; i < len; ++i)  
    {  
        if (i >= na) 
        {
            ax[i] = 0;
            ay[i] = 0;
        }  
        else
        {
            ax[i] = a[i];
            ay[i] = 0;
        }  
    }  
    fft(ax, ay, len, 0);  
    for (int i = 0; i < len; ++i)  
    {  
        if (i >= nb)
        {
            bx[i] = 0;
            by[i] = 0;
        }  
        else 
        {
            bx[i] = b[i];
            by[i] = 0;
        }  
    }  
    fft(bx, by, len, 0);  
    for (int i = 0; i < len; ++i)  
    {  
        double cx = ax[i] * bx[i] - ay[i] * by[i];  
        double cy = ax[i] * by[i] + ay[i] * bx[i];  
        ax[i] = cx;
        ay[i] = cy;  
    }  
    fft(ax, ay, len, 1);  
    for (int i = 0; i < len; ++i)  
        ans[i] = (int)(ax[i] + 0.5);  
    return len;  
}  

string mul(string sa, string sb)  
{  
    int l1, l2, l;  
    int i;  
    string ans;  
    memset(sum, 0, sizeof(sum));  
    l1 = sa.size();  
    l2 = sb.size();  
    for (i = 0; i < l1; i++)  
        x1[i] = sa[l1 - i - 1] - '0';  
    for (i = 0; i < l2; i++)  
        x2[i] = sb[l2 - i - 1] - '0';  
    l = solve(x1, l1, x2, l2, sum);  
    for (i = 0; i < l || sum[i] >= 10; i++)   
    {  
        sum[i + 1] += sum[i] / 10;  
        sum[i] %= 10;  
    }  
    l = i;  
    while(sum[l] <= 0 && l>0)    
        l--; 
    for(i = l; i >= 0; i--)    
        ans += sum[i] + '0';   
    return ans;  
}  

string Pow(string a, int n)  
{  
    if(n == 1) return a;  
    if(n & 1) return mul(Pow(a, n - 1), a);  
    string ans = Pow(a, n / 2);  
    return mul(ans, ans);  
}  

int main()  
{  
    ios::sync_with_stdio(false);
    cin.tie(0);    
    while (cin >> a >> b) 
        cout << Pow(a, b) << endl;  
    return 0;  
}  

11.高精度GCD

復雜度玄學

#include <iostream>
#include <cstdio>  
#include <cstring>  
#include <algorithm>
  
using namespace std;  
//#define DEBUG(x) cerr << #x << "=" << x << endl
const int L = 110;

string a, b;
  
string add(string a, string b)  
{  
    string ans;  
    int na[L] = {0};
    int nb[L] = {0};  
    int la = a.size();
    int lb = b.size();  
    for (int i = 0; i < la; i++) 
        na[la - 1 - i] = a[i] - '0';  
    for (int i = 0; i < lb; i++)
        nb[lb - 1 - i] = b[i] - '0';  
    int lmax = la > lb ? la : lb;  
    for (int i = 0; i < lmax; i++) 
    {
        na[i] += nb[i];
        na[i + 1] += na[i] / 10;
        na[i] %= 10;
    }  
    if (na[lmax]) 
        lmax++;  
    for (int i = lmax - 1; i >= 0; i--) 
        ans += na[i] + '0';  
    return ans;  
}  

string mul(string a, string b)  
{  
    string s;  
    int na[L];
    int nb[L];
    int nc[L];
    int La = a.size();
    int Lb = b.size();  
    fill(na, na + L, 0);
    fill(nb, nb + L, 0);
    fill(nc, nc + L, 0);  
    for (int i = La - 1; i >= 0; i--)
        na[La - i] = a[i] - '0';  
    for (int i = Lb - 1; i >= 0; i--) 
        nb[Lb - i] = b[i] - '0';  
    for (int i = 1; i <= La; i++)  
        for (int j = 1; j <= Lb; j++)  
            nc[i + j - 1] += na[i] * nb[j];  
    for (int i = 1; i <= La + Lb; i++)  
    {
        nc[i + 1] += nc[i] / 10;
        nc[i] %= 10;
    } 
    if (nc[La + Lb]) 
        s += nc[La + Lb] + '0';  
    for (int i = La + Lb - 1; i >= 1; i--)  
        s += nc[i] + '0'; 
    return s;  
}  

int sub(int *a, int *b, int La, int Lb)  
{  
    if (La < Lb) return -1;  
    if (La == Lb)  
    {  
        for (int i = La - 1; i >= 0; i--)  
            if (a[i] > b[i]) break;  
                else if(a[i] < b[i]) return -1; 
    }  
    for (int i = 0; i < La; i++) 
    {  
        a[i] -= b[i];  
        if (a[i] < 0) 
        {
            a[i] += 10;
            a[i + 1]--;
        }  
    }  
    for (int i = La - 1; i >= 0; i--)  
        if (a[i]) return i + 1;  
    return 0;
}  

string div(string n1, string n2, int nn)  
{  
    string s;
    string v;
    int a[L];
    int b[L];
    int r[L];
    int La = n1.size();
    int Lb = n2.size();
    int i, tp = La;  
    fill(a, a + L, 0);
    fill(b, b + L, 0);
    fill(r, r + L, 0); 
    for (i = La - 1; i >= 0; i--) 
        a[La - 1 - i] = n1[i] - '0';  
    for (i = Lb - 1; i >= 0; i--)
        b[Lb - 1 - i] = n2[i] - '0';  
     if (La < Lb || (La == Lb && n1 < n2)) 
     {  
            return n1;
     }
     int t = La - Lb;
     for (int i = La - 1; i >= 0; i--) 
         if (i >= t) 
             b[i] = b[i - t];  
             else b[i] = 0;  
     Lb = La;  
     for (int j = 0; j <= t; j++)  
     {  
         int temp;  
         while ((temp = sub(a, b + j, La, Lb - j)) >= 0)  
         {  
             La = temp;  
             r[t - j]++;  
         }  
     }  
     for (i = 0; i < L - 10; i++) 
     {
          r[i + 1] += r[i] / 10;
         r[i] %= 10;
     } 
     while (!r[i]) i--;  
     while (i >= 0) 
          s += r[i--] + '0';  
     i = tp;  
     while (!a[i]) i--;  
     while (i >= 0) 
          v += a[i--] + '0';  
     if (v.empty()) v = "0";  
     if(nn == 1) return s;  
     if(nn == 2) return v;  
}  

bool judge(string s)  
{  
    for (int i = 0; i < s.size(); i++)  
        if (s[i] != '0') 
            return false;  
    return true;  
}  

string gcd(string a, string b)  
{  
    string t;  
    while (!judge(b)) 
    {  
        t = a;  
        a = b;  
        b = div(t, b, 2); 
    }  
    return a;  
}  

int main()  
{  
    ios::sync_with_stdio(false); 
    cin.tie(0);   
    while (cin>>a>>b) 
        cout<<gcd(a,b)<<endl;  
    return 0;  
}  

12.高精度進制轉換

復雜度O(n*n)

#include <iostream>
#include <cstdio>
#include <cstring>  
#include <algorithm>
  
using namespace std; 
//#define DEBUG(x) cerr << #x << "=" << x << endl  

string s;

bool judge(string s)  
{  
    for (int i = 0; i < s.size(); i++)  
        if (s[i] != '0') 
            return 1;  
    return 0;  
}  

string solve(string s, int n, int m) 
{  
    string r, ans;  
    int d = 0;  
    if (!judge(s)) return "0";
    while (judge(s)) 
    {  
        for (int i = 0; i < s.size(); i++)  
        {  
            r += (d * n + s[i] - '0') / m + '0'; 
            d = (d * n + (s[i] - '0')) % m; 
        }  
       s = r;
       r = ""; 
       ans += d + '0'; 
       d = 0;
    }  
    reverse(ans.begin(), ans.end());      
    return ans;  
}  

int main()  
{  
    ios::sync_with_stdio(false);
    cin.tie(0);  
    while(cin>>s)  
    {  
        cout << solve(s, 10, 7) << endl;  
    }  
    return 0;  
}  

13.高精度平方根

復雜度O(n*n*n)

#include <iostream>  
#include <cstring>  
#include <cstdio>  
#include <algorithm> 
 
using namespace std;  
//#define DEBUG(x) cerr << #x << "=" << x << endl
const int L = 2018;  

string n;  
int t; 

string add(string a, string b)//只限兩個非負整數相加  
{  
    string ans;  
    int na[L] = {0};
    int nb[L] = {0};  
    int la = a.size();
    int lb = b.size();  
    for (int i = 0; i < la; i++) 
        na[la - 1 - i] = a[i] - '0';  
    for (int i = 0; i <lb; i++) 
        nb[lb - 1 - i] = b[i] - '0';  
    int lmax = la > lb ? la : lb;  
    for (int i = 0; i < lmax; i++) 
    {
        na[i] += nb[i];
        na[i + 1] += na[i] / 10;
        na[i] %= 10;
    }  
    if (na[lmax]) lmax++;  
    for (int i = lmax - 1; i >= 0; i--) 
        ans += na[i] + '0';  
    return ans;  
}  

string sub(string a, string b)//只限大的非負整數減小的非負整數  
{  
    string ans;  
    int na[L] = {0};
    int nb[L] = {0};  
    int la = a.size();
    int lb = b.size();  
    for (int i = 0; i < la; i++) 
        na[la - 1 - i] = a[i] - '0';  
    for (int i = 0; i < lb; i++) 
        nb[lb - 1 - i] = b[i] - '0';  
    int lmax = la > lb ? la : lb;  
    for (int i = 0; i < lmax; i++)  
    {  
        na[i] -= nb[i];  
        if (na[i] < 0)
        {
            na[i] += 10;
            na[i + 1]--;
        }  
    }  
    while (!na[--lmax] && lmax > 0) lmax++;  
    for (int i = lmax - 1; i >= 0; i--) 
        ans += na[i] + '0';  
    return ans;  
}  

string mul(string a, string b)//高精度乘法a,b,均為非負整數  
{  
    string s;  
    int na[L];
    int nb[L];
    int nc[L];
    int La = a.size();
    int Lb = b.size(); 
    fill(na, na + L, 0);
    fill(nb, nb + L, 0);
    fill(nc, nc + L, 0);  
    for (int i = La - 1; i >= 0; i--) 
        na[La - i] = a[i] - '0';
    for (int i = Lb - 1; i >= 0; i--) 
        nb[Lb - i] = b[i] - '0';  
    for (int i = 1; i <= La; i++)  
        for (int j = 1; j <= Lb; j++)  
            nc[i + j - 1] += na[i] * nb[j];  
    for (int i = 1; i <= La + Lb; i++)  
    {
        nc[i + 1] += nc[i] / 10;
        nc[i] %= 10;
    } 
    if (nc[La + Lb]) 
        s += nc[La + Lb] + '0'; 
    for (int i = La + Lb - 1; i >= 1; i--)  
        s += nc[i] + '0';
    return s;  
}  

int sub(int *a, int *b, int La, int Lb)  
{  
    if (La < Lb) return -1;
    if (La == Lb)  
    {  
        for (int i = La - 1; i >= 0; i--)  
            if (a[i] > b[i]) break;  
                else if (a[i] < b[i]) return -1; 
    }  
    for (int i = 0; i < La; i++)
    {  
        a[i] -= b[i];  
        if (a[i] < 0) 
        {
            a[i] += 10;
            a[i + 1]--;
        }  
    }  
    for (int i = La - 1; i >= 0; i--)  
        if (a[i]) return i + 1;  
    return 0;
}  

string div(string n1, string n2, int nn)  
{  
    string s;
    string v;
    int a[L];
    int b[L];
    int r[L];
    int La = n1.size();
    int Lb = n2.size();
    int i, tp = La;  
    fill(a, a + L, 0);
    fill(b, b + L, 0);
    fill(r, r + L, 0); 
    for (i = La - 1; i >= 0; i--) 
        a[La - 1 - i] = n1[i] - '0';  
    for (i = Lb - 1; i >= 0; i--)
        b[Lb - 1 - i] = n2[i] - '0';  
    if (La < Lb || (La == Lb && n1 < n2)) 
    {  
           return n1;
    }
    int t = La - Lb;
    for (int i = La - 1; i >= 0; i--) 
        if (i >= t) 
            b[i] = b[i - t];  
        else b[i] = 0;  
    Lb = La;  
    for (int j = 0; j <= t; j++)  
    {  
       int temp;  
       while ((temp = sub(a, b + j, La, Lb - j)) >= 0)  
        {  
            La = temp;  
            r[t - j]++;  
        }  
    }  
    for (i = 0; i < L - 10; i++) 
    {
         r[i + 1] += r[i] / 10;
         r[i] %= 10;
    } 
    while (!r[i]) i--;  
    while (i >= 0) 
         s += r[i--] + '0';  
    i = tp;  
    while (!a[i]) i--;  
    while (i >= 0) 
         v += a[i--] + '0';  
    if (v.empty()) v = "0";  
    if (nn == 1) return s;  
    if (nn == 2) return v;  
}  

bool cmp(string a, string b)  
{  
    if (a.size() < b.size()) return 1; 
    if (a.size() == b.size() && a <= b) return 1;  
    return 0;  
}  

string BigInterSqrt(string n)  
{  
    string l = "1";
    string r = n;
    string mid;
    string ans;  
    while (cmp(l, r))  
    {  
        mid = div(add(l, r), "2", 1);  
        if (cmp(mul(mid, mid), n))
        {
            ans = mid;
            l = add(mid, "1");
        }  
        else r = sub(mid, "1");  
    }  
    return ans;  
}  

string DeletePreZero(string s)  
{  
    int i;  
    for (i = 0; i < s.size(); i++)  
        if (s[i] != '0') break;  
    return s.substr(i);  
}  

int main()  
{   
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> t;  
    while (t--)  
    {  
        cin >> n;  
        n = DeletePreZero(n);  
        cout << BigInterSqrt(n) << endl;    
    }  
    return 0;  
}