dp的开关路灯

An Easy Game

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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One day, Edward and Flandre play a game. Flandre will show two 01-strings s₁ and s₂, the lengths of two strings are n. Then, Edward must move exact k steps. In each step, Edward should change exact m positions of s₁. That means exact m positions of s₁, '0' will be changed to '1' and '1' will be changed to '0'.

The problem comes, how many different ways can Edward change s₁ to s₂ after k steps? Please calculate the number of the ways mod 1000000009.

Input

Input will consist of multiple test cases and each case will consist of three lines. The first line of each case consist of three integers n (1 ≤ n ≤ 100), k (0 ≤ k ≤ 100), m (0 ≤ m ≤ n). The second line of each case is a 01-string s₁. The third line of each case is a 01-string s₂.

Output

For each test case, you should output a line consist of the result.

Sample Input

3 2 1100001

Sample Output

2

Hint

100->101->001100->000->001

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const long long int  MOD = 1000000009LL;long long  C[110][110],dp[110][110];int n,m,K;char s1[110],s2[110];int main(){    for(int i=0;i<110;i++) C[i][i]=1LL,C[i][0]=1LL;    for(int i=2;i<110;i++) for(int j=1;j<i;j++) C[i][j]=(C[i-1][j]+C[i-1][j-1])%MOD;    while(scanf("%d%d%d",&n,&m,&K)!=EOF)    {        memset(dp,0,sizeof(dp));        scanf("%s%s",s1,s2);        int nt=0;        for(int i=0;i<n;i++)            if(s1[i]!=s2[i]) nt++;        dp[0][nt]=1;        for(int i=1;i<=m;i++)        {            for(int j=0;j<=n;j++)            {                for(int k=max(0,j-K);k<=min(n,j+K);k++)                {                    ///   dp[i][j]+=dp[i-1][k]*C[][]*C[][];                    int deta=j-k;///不同的位置个数变化量                    if(deta>=0)                    {                        if(deta==K||(K-deta)%2==0)                        {                            dp[i][j]=(dp[i][j]+((dp[i-1][k]*C[k][(K-deta)/2])%MOD*C[n-k][deta+(K-deta)/2])%MOD)%MOD;                        }                    }                    else if(deta<0)                    {                        deta*=-1;                        if(deta==K||(K-deta)%2==0)                        {                            dp[i][j]=(dp[i][j]+((dp[i-1][k]*C[k][deta+(K-deta)/2])%MOD*C[n-k][(K-deta)/2])%MOD)%MOD;                        }                    }                }            }        }        printf("%lld\n",dp[m][0]);    }    return 0;}