HDU 5800 To My Girlfriend

背包变形。dp[i][j][g][h]表示前i个数字，和为j，有g个必选，有h个必不选的方案数。

答案为sum{dp[n][j][2][2]}*4

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
inline int read()
{
    char c = getchar();  while(!isdigit(c)) c = getchar();
    int x = ;
    while(isdigit(c)) { x = x *  + c - ''; c = getchar(); }
    return x;
}

const int maxn=+;
int dp[maxn][maxn][][],f[maxn][maxn][][];
int mod=1e9+;
int T,n,s,a[maxn];

int main()
{
    scanf("%d",&T); while(T--)
    {
        scanf("%d%d",&n,&s);
        for(int i=;i<=n;i++) scanf("%d",&a[i]);
        memset(dp,,sizeof dp); dp[][][][]=;

        for(int i=;i<=n;i++)
        {
            for(int j=;j<=s;j++)
            {
                for(int p=;p<=;p++)
                {
                    for(int h=;h<=;h++)
                    {
                        int num=;
                        num=(num+dp[i-][j][p][h])%mod;
                        if(j-a[i]>=) num=(num+dp[i-][j-a[i]][p][h])%mod;
                        if(p->=&&j>=a[i]) num=(num+dp[i-][j-a[i]][p-][h])%mod;
                        if(h->=) num=(num+dp[i-][j][p][h-])%mod;
                        dp[i][j][p][h]=num;
                    }
                }
            }
        }
        LL ans=;
        for(int i=;i<=s;i++) ans=(ans+dp[n][i][][])%mod;
        printf("%lld\n",(ans*)%(LL)mod);
    }
    return ;
}