hdu5456 Matches Puzzle Game（记忆化dfs+dp）

A - B = C 可以转化成 B + C = A

题解：http://m.blog.csdn.net/blog/stl112514/48678307

http://acm.hdu.edu.cn/showproblem.php?pid=5456

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
int n,m;
int a[10]={6,2,5,5,4,5,6,3,7,6};
ll dp[510][2][2][2];
int dfs(int num,int cr,int b,int c){
	if(num>n)
		return 0;		
	if(dp[num][cr][b][c]!=-1)
		return dp[num][cr][b][c];
	if(b==1&&c==1){ //两个加数都已经到最高位 
		if(num==n&&cr==0)  //没有进位 
			return 1;
		if(num+a[1]==n&&cr==1) //有进位，剩余的火柴数刚好等于进位"1" 
			return 1;
		return 0;
	}		
	if(num==n)   //火柴已经用完，又不符合上面的判别条件 
		return 0;
	dp[num][cr][b][c]=0;
	if(b==0){
		if(c==0){
			for(int i=0;i<10;i++){ 
				for(int j=0;j<10;j++){
					int t=a[i]+a[j]+a[(i+j+cr)%10];
					dp[num][cr][b][c]+=dfs(num+t,(i+j+cr)/10,0,0);
					if(i!=0)
						dp[num][cr][b][c]+=dfs(num+t,(i+j+cr)/10,1,0);
					if(j!=0)
						dp[num][cr][b][c]+=dfs(num+t,(i+j+cr)/10,0,1);
					if(i!=0&&j!=0)
						dp[num][cr][b][c]+=dfs(num+t,(i+j+cr)/10,1,1);
					dp[num][cr][b][c]%=m;
				}
			} 
		}
		else{
			for(int i=0;i<10;i++){
				int t=a[i]+a[(i+cr)%10];
				dp[num][cr][b][c]+=dfs(num+t,(i+cr)/10,0,1);
				if(i!=0)
					dp[num][cr][b][c]+=dfs(num+t,(i+cr)/10,1,1);
			}
		}
	}
	else{
		if(c==0){
			for(int i=0;i<10;i++){
				int t=a[i]+a[(i+cr)%10];
				dp[num][cr][b][c]+=dfs(num+t,(i+cr)/10,1,0);
				if(i!=0)
					dp[num][cr][b][c]+=dfs(num+t,(i+cr)/10,1,1);
			}
		}
	}
	return dp[num][cr][b][c]%=m;
}
int main(){
	int T;
	scanf("%d",&T);
	for(int cs=1;cs<=T;cs++){
		cin>>n>>m;
		n-=3;
		memset(dp,-1,sizeof(dp));
		printf("Case #%d: %d\n",cs,dfs(0,0,0,0));
	}
}

#include<iostream>
#include<algorithm>
#include<string>
#include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};
#include<set>//int gcd(int a,int b){return b?gcd(b,a%b):a;}
#include<vector>
#include<cmath>
#include<queue>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define mod 1e9+7
#define ll long long
using namespace std;
int n,m;
ll dp[505][2][2][2];  //不知道为什么int不行。。 
int x[10]={6,2,5,5,4,5,6,3,7,6};  
int dfs(int num,int f,int a,int b){
    if(num>n)
        return 0;
    if(dp[num][f][a][b]!=-1)
        return dp[num][f][a][b];
    if(num==n){
        if(f==0&&a==1&&b==1)
            return 1;
        return 0;
    }
    if(a==1&&b==1){
        if(f==1&&num+2==n)
            return 1;
        return 0;
    }
    dp[num][f][a][b]=0; 
    if(a==0&&b==0){
        for(int i=0;i<10;++i){
            for(int j=0;j<10;++j){
                dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[j]+x[(i+j+f)%10],(i+j+f)/10,0,0))%m;
                if(i!=0)
                    dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[j]+x[(i+j+f)%10],(i+j+f)/10,1,0))%m;
                if(j!=0)
                    dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[j]+x[(i+j+f)%10],(i+j+f)/10,0,1))%m;
                if(i!=0&&j!=0)
                    dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[j]+x[(i+j+f)%10],(i+j+f)/10,1,1))%m;
            }
        }
    }
    else if(a==1&&b==0){
        for(int j=0;j<10;++j){
            dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[j]+x[(j+f)%10],(j+f)/10,1,0))%m;
            if(j!=0)
                dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[j]+x[(j+f)%10],(j+f)/10,1,1))%m;
        }
    }
    else if(a==0&&b==1){
        for(int i=0;i<10;++i){
            dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[(i+f)%10],(i+f)/10,0,1))%m;
            if(i!=0)
                dp[num][f][a][b]=(dp[num][f][a][b]+dfs(num+x[i]+x[(i+f)%10],(i+f)/10,1,1))%m;
        }
    }
    return dp[num][f][a][b];
}
int main(){
    int t,cnt=0;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        n-=3;
        memset(dp,-1,sizeof(dp));
        printf("Case #%d: %d\n",++cnt,dfs(0,0,0,0));
    }
    return 0;
}