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;
}