ZOJ 3791 An Easy Game

最新推荐文章于 2020-01-27 10:07:07 发布

原创最新推荐文章于 2020-01-27 10:07:07 发布 · 1.3k 阅读

0 ·

CC 4.0 BY-SA版权

DP 专栏收录该内容

150 篇文章

订阅专栏

本文讨论了一种使用动态规划（DP）解决特定游戏变换问题的方法。通过改变字符串中指定数量的位置，目标是将一个字符串转换为另一个字符串。文章详细解释了如何通过动态规划计算在给定步骤下实现目标字符串的不同方式的数量，同时考虑到字符串长度、步骤限制和位置更改的约束。

这种数据范围一看就想到DP....

dp[i][j]:在第i轮有j个不同位置的可能的方法数，配合组合数搞一搞就可以了。。。注意超INT

An Easy Game

Time Limit: 2 Seconds Memory Limit: 65536 KB

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 1
100
001

Sample Output

Hint

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

Author: CHEN, Zemin
Source: ZOJ Monthly, June 2014

#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;
}