2018农大校赛（dp大法）

0&1

Description

Input

第一行是一个不超过100的整数t，代表了样例的组数

接下来有多行，每行两个整数n和k，

Output

对于每组输入样例，先输出它的序号标识(‘CASE 1:’, ‘CASE 2:’, etc).

再输出其满足条件的数的个数

Sample Input 1 

5
6 3
6 4
6 2
26 3
64 2

Sample Output 1

Case 1: 1
Case 2: 3
Case 3: 6
Case 4: 1662453
Case 5: 465428353255261088

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define ll long long
#define ull unsigned long long 
using namespace std;
ll dp[66][66][105]; 
void dis(int a[], int n){
	printf("总数为%d个\n",n); 
	for(int i = 0; i < n; i++) 	cout<<a[i]<<", ";
	cout<<endl<<"------------------"<<endl;		
}

const int mx = 0;
int main(){
	int T,n,mod;
	scanf("%d",&T);
	//T = 10;
	for(int ca = 1; ca <= T; ca++){
		scanf("%d%d",&n,&mod);
		memset(dp,0,sizeof(dp));
		if(n%2||n== 0||mod==0){
			printf("Case %d: %lld\n",ca,0);
			continue;	
		}
	/*	if(k== 0){
			printf("Case %d: %lld\n",0);
			continue;	
		}*/
		dp[1][1][1%mod] = 1; 
		
		for(int i = 2; i <=n; i++){
			for(int j = 0; j <= i; j++){
				for(int k = 0; k < mod; k++){
					if(i >0&&j>0)
						dp[i][j][(k*2+1)%mod] += dp[i-1][j-1][k];   //加1的 
					if(i >0)
						dp[i][j][(k*2)%mod] += dp[i-1][j][k];
				}
			}
		}
	//	cout<<dp[2][1][0]<<endl;
	//	ll ans = 0;
		printf("Case %d: %lld\n",ca,dp[n][n/2][0]);
		//cout<<dp[n][n/2][0]<<endl;
	}
	
	
	return 0;
}