# Zoj?#an easy game

Source: Internet
Author: User
Tags first string

Two equal-length strings are given. Each time M numbers are required to be changed, K numbers must be changed each time, and the number of solutions from the first string to the second string must be calculated.

DP. F [I] [J] Number of solutions with J locations changed after step I is changed. Then, we can directly enumerate the positions that are currently changed, which are overlapped.

Then output the statistical answer.

`#include <iostream>#include <cstring>#include <cstdio>#define M 1000000009#define maxn 105typedef long long ll;using namespace std;ll C[maxn][maxn];ll f[maxn][maxn];int n,k,m,change;ll ans;ll power(ll A,ll B){    ll tot=1;    while (B){        if (B&1) tot=tot*A%M;        A=A*A%M,B>>=1;    }    return tot;}void _init(){    memset(C,0,sizeof C);    C=1;    for (int i=1; i<maxn; i++){        C[i]=1;        for (int j=1; j<=i; j++) C[i][j]=(C[i-1][j]+C[i-1][j-1])%M;    }}int main(){    _init();    char s1[maxn],s2[maxn];    while (scanf("%d%d%d",&n,&k,&m)!=EOF){        change=0;        scanf("%s%s",s1,s2);        for (int i=0; i<n; i++)            if (s1[i]!=s2[i]) change++;        memset(f,0,sizeof f);        f=1;        for (int i=0; i<k; i++)//after the ith time of changes            for (int j=0; j<=n; j++){//the number of 1 is j                if (f[i][j]==0) continue;                for (int x=max(0,j+m-n); x<=min(j,m); x++){                    f[i+1][j-x+m-x]+=f[i][j]*(C[j][x]*C[n-j][m-x]%M)%M;                    f[i+1][j-x+m-x]%=M;                }            }        ans=f[k][change]*power(C[n][change],M-2)%M;        printf("%d\n",(int)ans);    }    return 0;}`

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