Codeforces Round #607 (Div. 2)

太久沒做題就會變得很菜。

題目鏈接：https://codeforces.com/contest/1281/

---

A：

白給。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

int t;
const int maxn=1010;
char s[maxn];

int main() {
    scanf("%d",&t);
    while(t--){
        scanf("%s",s+1);
        int len=strlen(s+1);
        if (s[len]=='o') puts("FILIPINO");
        else if (s[len]=='u') puts("JAPANESE");
        else puts("KOREAN");
    }
    return 0;
}

B：

顯然對於每一位，只能跟后面的字符交換。如果只是O(n)掃，遇到第一個s[i]>c[i]的情況才考慮替換的話是錯的。應該先把s的副本sort一遍，這樣得到的結果是s每個位置交換能得到的最小的字符。當s[i]>s_copy[i]，即當前字符有優化空間時就立即考慮替換。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (minpos<<1)
#define rson (minpos<<1|1)
/* namespace */
using namespace std;
/* header end */

int t;
string a,b;

int main() {
    cin>>t;
    while (t--){
        cin>>a>>b;
        if (a<b){
            cout<<a<<'\n';
            continue;
        }
        string s=a;
        sort(s.begin(),s.end());
        int flag=0;
        for (int i=0;i<(int)a.size();i++){
            // current char is not at the best position
            if (a[i]>s[i]){
                // it can swap with the back char only
                for (int j=i+1;j<(int)a.size();j++){
                    swap(a[i],a[j]);
                    if (a<b){
                        cout<<a<<'\n';
                        flag=1;
                        i=(int)a.size();
                        break;
                    }
                    swap(a[i],a[j]);
                }
            }
        }
        if (!flag) puts("---");
    }
    return 0;
}

C：

看似很數學，其實直接模擬就過了。注意當前字符串長度大於等於x時則停止延長字符串即可。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int mod = 1e9 + 7;
int t;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> t;
    while (t--) {
        int x; string s; cin >> x >> s;
        int currLen = (int)s.size();
        for (int i = 0; i < x; i++) {
            int curr = s[i] - '0';
            // length need to grow
            if (currLen < x) {
                for (int j = 1; j < curr; j++)
                    if ((int)s.size() < x) {
                        for (int l = i + 1; l < currLen; l++)
                            if ((int)s.size() < x) s += s[l];
                    }
            }
            // calculate current length of string
            int d = (currLen - i - 1 + mod) % mod;
            currLen = ((i + 1) + (ll)d * curr) % mod;
        }
        cout << currLen << '\n';
    }
    return 0;
}

D：

分類討論題。很顯然答案只能在[0,4]之間取值。枚舉每一行每一列，對於每一行，檢查其是否為全A、A在邊緣和A在內部的情況，維護答案。列同理。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 100;
int t, n, m;
char a[maxn][maxn];

int main() {
    scanf("%d", &t);
    while (t--) {
        int foundA = 0, foundP = 0, ans = 4;
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++) {
            scanf("%s", a[i] + 1);
            for (int j = 1; j <= m; j++)
                if (a[i][j] == 'A') foundA = 1;
                else foundP = 1;
        }
        if (!foundA) {
            puts("MORTAL");
            continue;
        }
        if (!foundP) {
            puts("0");
            continue;
        }
        for (int i = 1; i <= n; i++) {
            int maxx = -1, minn = 1000;
            for (int j = 1; j <= m; j++) {
                maxx = max(maxx, (int)a[i][j]), minn = min(minn, (int)a[i][j]);
                if (a[i][j] == 'A') {
                    int t = 4;
                    if (i == 1 || i == n) t--;
                    if (j == 1 || j == m) t--;
                    ans = min(ans, t);
                }
            }
            if (maxx == minn && maxx == 'A') {
                if (i == 1 || i == n) ans = min(ans, 1);
                else ans = min(ans, 2);
            }
        }
        for (int j = 1; j <= m; j++) {
            int maxx = -1, minn = 1000;
            for (int i = 1; i <= n; i++) {
                maxx = max(maxx, (int)a[i][j]), minn = min(minn, (int)a[i][j]);
            }
            if (maxx == minn && maxx == 'A') {
                if (j == 1 || j == m) ans = min(ans, 1);
                else ans = min(ans, 2);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

E：

待補。