1 //數據量較小,1000 2 void C() 3 { 4 c[0][0]=1; 5 for(int i=1;i<=10;i++) 6 { 7 for(int j=0;j<=i;j++){ 8 if(j==0||j==i) c[i][j]=1; 9 else{ 10 c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod; 11 } 12 printf("%d%c",c[i][j],j==i?'\n':' '); 13 } 14 printf("\n"); 15 } 16 } 55 // 0≤n≤1e18，0≤m≤1e6，1≤mod≤1e9，用盧卡斯定理,mod為素數 56 void egcd(ll a,ll b,ll &x,ll &y) 57 { 58 ll d; 59 if(!b) { 60 x=1,y=0; 61 return ; 62 } 63 else{ 64 egcd(b,a%b,y,x); 65 y-=(a/b)*x; 66 } 67 } 68 ll inv(ll n) 69 { 70 ll x,y; 71 egcd(n,mod,x,y); 72 return (x%mod+mod)%mod; 73 } 74 ll C(ll n,ll m) 75 { 76 if(m>n) return 0; 77 ll ans=1ll; 78 for(ll i=1;i<=m;i++) 79 { 80 ans=(ans*(n-m+i)%mod*inv(i)%mod)%mod; 81 } 82 return ans; 83 }
C(n, m) % p = C(n / p, m / p) * C(n%p, m%p) % p 84 ll solve(ll n,ll m) 85 { 86 if(m==0) return 1; 87 return C(n%mod,m%mod)*solve(n/mod,m/mod)%mod; 88 } 89 90 91 //0≤m≤n≤1e18，1≤mod≤1e6，用盧卡斯定理，mod為素數 92 ll f[1000009]; 93 void init() 94 { 95 f[0]=1; 96 for(ll i=1;i<=mod;i++) 97 { 98 f[i]=(f[i-1]*i)%mod; 99 } 100 return ; 101 } 102 void egcd(ll a,ll b,ll &x,ll &y) 103 { 104 ll d; 105 if(!b){ 106 x=1,y=0; 107 return ; 108 } 109 else{ 110 egcd(b,a%b,y,x); 111 y-=(a/b)*x; 112 } 113 } 114 ll inv(ll n) 115 { 116 ll x,y; 117 egcd(n,mod,x,y); 118 return (x%mod+mod)%mod; 119 } 120 ll C(ll n,ll m) 121 { 122 ll ans=1ll; 123 while(n&&m){ 124 ll nn=n%mod,mm=m%mod; 125 if(nn<mm) return 0; 126 ans=(ans*f[nn]%mod*inv(f[mm]*f[nn-mm]%mod)%mod)%mod; 127 n/=mod; 128 m/=mod; 129 } 130 return ans; 131 }
1 //0<=n,mod<=1e6. 2 ll fac[N] = {1, 1}, inv[N] = {1, 1}, f[N] = {1, 1}; 3 ll C(ll a, ll b) { 4 if (b > a)return 0; 5 return fac[a] * inv[b] % mod * inv[a - b] % mod; 6 } 7 void init() { //快速計算階乘的逆元，mod為素數 8 for (int i = 2; i < N; i++) { 9 fac[i] = fac[i - 1] * i % mod; 10 f[i] = (mod - mod / i) * f[mod % i] % mod; 11 inv[i] = inv[i - 1] * f[i] % mod; 12 } 13 }
1 //c(n,m)%M ：M為p[i]的乘積 2 //n,m,M都在10^18范圍 3 //p[i]都是素數，且都小於10^5 4 // HDU 5446 5 #include <iostream> 6 #include <cstdio> 7 #include <cstring> 8 #include <vector> 9 #include <queue> 10 using namespace std; 11 typedef long long ll; 12 ll p[15],an[15]; 13 ll fac[100005],inv[100005]; 14 ll pow_mod(ll a, ll n, ll mod) 15 { 16 ll ret = 1; 17 while (n) 18 { 19 if (n&1) ret = ret * a % mod; 20 a = a * a % mod; 21 n >>= 1; 22 } 23 return ret; 24 } 25 void ini(int x) 26 { 27 fac[0] = 1; 28 for(int i = 1; i < x; i++) fac[i] = fac[i-1]*i%x; 29 inv[x - 1] = pow_mod(fac[x-1],x-2,x); 30 for(int i = x - 2; i >= 0; i--) inv[i] = inv[i+1] * (i+1) % x; 31 } 32 ll c(ll a,ll b,ll p) 33 { 34 if(a < b || a < 0 || b < 0) 35 return 0; 36 return fac[a]*inv[b]%p*inv[a-b]%p; 37 } 38 ll lucas(ll a,ll b, ll p) 39 { 40 if( b == 0) 41 return 1; 42 return lucas(a/p,b/p,p)*c(a%p,b%p,p)%p; 43 } 44 ll ex_gcd(ll a, ll b, ll& x, ll& y) 45 { 46 if (b == 0) 47 { 48 x = 1; 49 y = 0; 50 return a; 51 } 52 ll d = ex_gcd(b, a % b, y, x); 53 y -= x * (a / b); 54 return d; 55 } 56 ll mul(ll a, ll b, ll mod) 57 { 58 a = (a % mod + mod) % mod; 59 b = (b % mod + mod) % mod; 60 ll ret = 0; 61 while(b) 62 { 63 if(b&1) 64 { 65 ret += a; 66 if(ret >= mod) ret -= mod; 67 } 68 b >>= 1; 69 a <<= 1; 70 if(a >= mod) a -= mod; 71 } 72 return ret; 73 } 74 ll china(ll n,ll a[],ll b[]) 75 { 76 ll M = 1,d,y,x= 0; 77 for(int i = 0; i < n; i++) 78 { 79 M *= b[i]; 80 } 81 for(int i = 0; i < n; i++) 82 { 83 ll w = M/b[i]; 84 ex_gcd(b[i],w,d,y); 85 x = (x + mul(mul(y, w, M), a[i], M));//可能超范圍 86 } 87 return (x+M) % M; 88 } 89 int main() 90 { 91 int t,k; 92 ll n,m; 93 scanf("%d",&t); 94 while(t--) 95 { 96 scanf("%lld%lld",&n,&m); 97 scanf("%d",&k); 98 for(int i = 0; i < k; i++) 99 { 100 scanf("%lld",&p[i]); 101 ini(p[i]); 102 an[i] = lucas(n,m,p[i]); 103 } 104 printf("%lld\n",china(k,an,p)); 105 } 106 return 0; 107 }
B: Interseting paths 時間限制: 1 s 內存限制: 128 MB 題目描述 給定一個n行m列的迷宮，行的編號從頂端到底部編號為1−>n，列的編號從左到右為1−>m。 現在你處於迷宮的左上角(1,1)處，現在你要到網格的(n,m)處。並且，你只能在當前的格子向右走或者向下走。即(x,y)−>(x+1,y)或者(x,y),(x,y+1)。但是，當你走到兩個小矩陣內時就不能再走，這兩個小矩陣分別是行a−>n,列1−>b。和行1−>c,列d,m.並且數據保證1≤c<a≤n,1≤b<d≤m. 現在，你想知道從(1,1)走到(n,m)有多少種方式。 輸入 第一行一個整數t，代表有t組測試數據 接下來t行每行6個整數n,m,a,b,c,d. 1≤t≤20. 3≤n,m≤105,1≤c<a≤n,1≤b<d≤m. 輸出 對於每組數據，輸出對109+7取模之后的結果. 樣例輸入 2 3 3 3 1 1 3 10 7 8 4 3 5 樣例輸出 4 2975 提示 第一個樣例解釋 走在黑色區域是無效的。 來源 Ocean_Star_T
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <utility> #include <vector> #include <map> #include <queue> #include <stack> #include <cstdlib> typedef long long ll; #define lowbit(x) (x&(-x)) #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 using namespace std; const ll mod=1e9+7; ll n,m,a,b,c,d; const int N=1e6+9;//大於2e5 ll mul(ll a,ll b){ a%=mod;b%=mod; return a*b%mod; } ll mul1(ll a,ll b,ll c){ a%=mod;b%=mod;c%=mod; return a*b%mod*c%mod; } ll fac[N]={1,1},inv[N]={1,1},f[N]={1,1}; void init() { for(int i=2;i<N;i++){ fac[i]=fac[i-1]*i%mod; f[i]=(mod-mod/i)*f[mod%i]%mod; inv[i]=inv[i-1]*f[i]%mod; } } ll C(ll a,ll b) { if(b>a) return 0; return fac[a]*inv[b]%mod*inv[a-b]%mod; } int main() { init(); int t; scanf("%d",&t); while(t--){ scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&a,&b,&c,&d); ll sum1=0,sum2=0;
// 上面兩幅圖加一起即可 for(int i=b+1;i<=d-1;i++){ ll x=C(c+i-2,c-1); ll y=C(n+m-c-i-1,n-c-1); ll z=mul(x,y); sum1=(sum1%mod+z)%mod; } for(int j=c+1;j<=a-1;j++){ ll x=C(j+b-2,b-1); ll y=C(n+m-b-j-1,m-b-1); ll z=mul(x,y); sum2=(sum2%mod+z)%mod; } ll sum=(sum1%mod+sum2%mod)%mod; printf("%lld\n",sum); } return 0; }
//The 2018 ACM-ICPC China JiangSu Provincial Programming Contest B Massage JSZKC feels so bored in the classroom that he wants to send massages to his girl friend. However, he can't move to his girl friend by himself. So he has to ask his classmates for help. The classroom is a table of size N \times MN×M. We'll consider the table rows numbered from top to bottom 11 through NN, and the columns numbered from left to right 11 through MM. Then we'll denote the cell in row xx and column yy as (x, y)(x,y). And every cell sits a student. Initially JSZKC sits on the cell (1, 1)(1,1) . And his girl friend sits on cell (n, m)(n,m). A message can go from cell (x, y)(x,y) to one of two cells (x + 1, y)(x+1,y) and (x, y + 1)(x,y+1). JSZKC doesn't want to trouble his classmates too much. So his classmates(not including his girl friend) may not take massages more than once. It's obvious that he can only send out two massages. Please help JSZKC find the number of ways in which the two massages can go from cell (1, 1)(1,1) to cell (n, m)(n,m). More formally, find the number of pairs of non-intersecting ways from cell (1, 1)(1,1) to cell (n, m)(n,m) modulo 10000000071000000007 . Two ways are called non-intersecting if they have exactly two common points — the starting point and the final point. Input Format The input file contains several test cases, each of them as described below. The first line of the input contains one integers NN (2 \le N,M \le 1000)(2≤N,M≤1000), giving the number of rows and columns in the classroom. There are no more than 100100 test cases. Output Format One line per case, an integer indicates the answer mod 10000000071000000007. 樣例輸入 復制 2 2 2 3 3 3 樣例輸出 復制 1 1 3 題目來源 The 2018 ACM-ICPC China JiangSu Provincial Programming Contest
1 (x1, y1) 到 (x2, y2) 的所有路徑方案數：x = x2-x1, y = y2-y1, C(x+y, x) 就是方案數。 2 3 /* 4 A1(1,2) A2(2,1) 5 B1(n-1,m) B2(n,m-1) 6 並且只能從A1到B1,從A2到B2.組合數問題。 7 由容斥知：要再減去有交點的路線。 8 分析： 9 如果兩條路線有交點，那么A1一定可以到B2，A2一定會到B1 10 */
1 ll n,m; 2 void egcd(ll a,ll b,ll &x,ll &y,ll &d) 3 { 4 if(!b) d=a,x=1,y=0; 5 else{ 6 egcd(b,a%b,y,x,d); 7 y-=x*(a/b); 8 } 9 } 10 ll inv(ll t){ 11 ll x,y,d; 12 egcd(t,mod,x,y,d); 13 return d==1?(x%mod+mod)%mod:-1; 14 } 15 ll C(ll n,ll m){ 16 if(n<m) return 0; 17 ll ans=1; 18 for(int i=1;i<=m;i++){ 19 ans=(ans*(n-m+i)%mod*inv(i)%mod)%mod; 20 } 21 return ans%mod; 22 } 23 ll mul(ll a,ll b){ 24 a%=mod,b%mod; 25 return a*b%mod; 26 } 27 ll mull(ll a,ll b,ll c) 28 { 29 a%=mod,b%=mod,c%=mod; 30 return a*b%mod*c%mod; 31 } 32 int main() 33 { 34 while(~scanf("%lld%lld",&n,&m)){ 35 ll ans1=C(n+m-4,n-2); 36 ll ans2=C(n+m-4,n-1); 37 ll ans3=C(n+m-4,m-1);//不能用n-3，因為n-3可能為負數，那么結果就是1了。 38 ll ans4=(mul(ans1,ans1)-mul(ans2,ans3)+mod)%mod; 39 printf("%lld\n",ans4); 40 } 41 return 0; 42 }