HDU5800-To My Girlfriend

To My Girlfriend

                                                                   Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                           Total Submission(s): 1397    Accepted Submission(s): 543

Problem Description

Dear Guo

I never forget the moment I met with you.You carefully asked me: "I have a very difficult problem. Can you teach me?".I replied with a smile, "of course"."I have n items, their weight was a[i]",you said,"Let's define f(i,j,k,l,m) to be the number of the subset of the weight of n items was m in total and has No.i and No.j items without No.k and No.l items.""And then," I asked.You said:"I want to know

∑i=1n∑j=1n∑k=1n∑l=1n∑m=1sf(i,j,k,l,m)(i,j,k,laredifferent)

Sincerely yours,
Liao

 

Input

The first line of input contains an integer T(T≤15) indicating the number of test cases.
Each case contains 2 integers n, s (4≤n≤1000,1≤s≤1000). The next line contains n numbers: a1,a2,…,an (1≤ai≤1000).

 

Output

Each case print the only number — the number of her would modulo 109+7 (both Liao and Guo like the number).

 

Sample Input

24 41 2 3 44 41 2 3 4

 

Sample Output

88

 

Author

UESTC

 

Source

2016 Multi-University Training Contest 6

 

Recommend

wange2014

 

题意：给定n个数，其中选定若干数，这若干数的权值和为m，且这些数中没有下标为i,j的数，有下标为k,l的数的集合个数。

解题思路：

dp[i][j][s1][s2]，代表的是前i个物品，总权值为j，已有s1个必选，s2必不选的方案数。那么对于当前一个状态，有四种转移状态：

1.选中当前的，增加权值，增加必选个数。

2.选择当前的，增加权值，不增加必选个数。

3.不选中当前的，不增加权值，增加不必选个数。

4.不选中当前的，不增加权值，不增加必选个数。

因为i，j可以互换，l，k也可以互换，故而最后方案数乘以4即为所求

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;const int mod = 1e9 + 7;int dp[1001][1001][3][3];int n, s, a[1001];int f(int a, int b){int sum = 0;while (b){if (b & 1) { sum += a; sum %= mod; }b >>= 1;a = (a*2) % mod;}return sum;}int main(){int t;scanf("%d", &t);while (t--){scanf("%d%d", &n, &s);for (int i = 1; i <= n; i++) scanf("%d", &a[i]);memset(dp, 0, sizeof dp);dp[1][a[1]][0][0] = 1;dp[1][a[1]][1][0] = 1;dp[1][0][0][0] = 1;dp[1][0][0][1] = 1;for (int i = 2; i <= n; i++){for (int j = 0; j <= s; j++){for (int k = 0; k <= 2; k++){dp[i][j][k][0] = (dp[i][j][k][0] + dp[i - 1][j][k][0]) % mod;dp[i][j][k][1] = (dp[i][j][k][1] + dp[i - 1][j][k][0]) % mod;dp[i][j][k][1] = (dp[i][j][k][1] + dp[i - 1][j][k][1]) % mod;dp[i][j][k][2] = (dp[i][j][k][2] + dp[i - 1][j][k][1]) % mod;dp[i][j][k][2] = (dp[i][j][k][2] + dp[i - 1][j][k][2]) % mod;}if (j + a[i] > s) continue;for (int k = 0; k <= 2; k++){dp[i][j + a[i]][0][k] = (dp[i][j + a[i]][0][k] + dp[i - 1][j][0][k]) % mod;dp[i][j + a[i]][1][k] = (dp[i][j + a[i]][1][k] + dp[i - 1][j][0][k]) % mod;dp[i][j + a[i]][1][k] = (dp[i][j + a[i]][1][k] + dp[i - 1][j][1][k]) % mod;dp[i][j + a[i]][2][k] = (dp[i][j + a[i]][2][k] + dp[i - 1][j][1][k]) % mod;dp[i][j + a[i]][2][k] = (dp[i][j + a[i]][2][k] + dp[i - 1][j][2][k]) % mod;}}}int ans = 0;for (int i = 1; i <= s; i++)ans = (ans + dp[n][i][2][2]) % mod;printf("%d\n", f(ans,4));}return 0;}