題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=6146

題意：~

解法：原題。。http://blog.csdn.net/y990041769/article/details/21243895

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 10010;
LL a[maxn], b[maxn];
const LL mod = 1e9+7;
int n;

int main()
{
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        b[1]=1LL;
        for(int i=2; i<=n; i++) b[i]=(b[i-1]*2LL%mod)%mod;
        a[1]=1LL;
        a[2]=6LL;
        for(int i=3; i<=n; i++){
            a[i]=(2LL*a[i-1]+1LL*b[i]+4LL*a[i-2])%mod;
        }
        long long sum = 4*a[n];
        for(int i=2; i<n; i++){
            sum += ((8LL*b[n-i]*a[i-1])%mod+(8LL*a[n-i]*b[i-1])%mod)%mod;
            sum %= mod;
        }
        if(n==1) sum=2;
        printf("%lld\n", sum);
    }
    return 0;
}

HDU 6146 Pokémon GO DP,計數