Zeros and Ones UVA - 12063

#include<bits/stdc++.h>
using namespace std;

#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
typedef long long ll;

int N,K;
ll dp[65][2][65][110];
int bit[65];
/*
又是一道数位dp,还是很懵逼
1.核心：状态压缩--->难点：怎样压缩表示
2.思想：树形的dp
*/

ll dfs(int pos,int v,int limit,int num1,ll sum){
    if(pos==-1){
        if((num1==N/2)&&(sum==0)){
                return 1;
        }
        return 0;
    }
    if(!limit&&dp[pos][v][num1][sum]!=-1){
        return dp[pos][v][num1][sum];
    }
    ll ans=0;
    int max_num=limit?bit[pos]:1;
    rep(i,0,max_num+1){
        if(!i)
            ans+=dfs(pos-1,0,limit&&i==max_num,num1,(sum*2)%K);
        else
            ans+=dfs(pos-1,1,limit&&i==max_num,num1+1,(sum*2+1)%K);
    }
    if(!limit)dp[pos][v][num1][sum]=ans;
    return ans;
}

ll solve(int n){
    memset(dp,-1,sizeof(dp));
    rep(i,0,n)bit[i]=1;
    return dfs(n-2,1,1,1,1ll);
}

int main(){
    int T;
    scanf("%d",&T);
    rep(kase,1,T+1){
        scanf("%d %d",&N,&K);
        if(N&1||!K){
            printf("Case %d: 0\n",kase);
            continue;
        }
        printf("Case %d: %lld\n",kase,solve(N));

    }
    return 0;
}