To My Girlfriend

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1326 Accepted Submission(s): 515

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

2
4 4
1 2 3 4
4 4
1 2 3 4

Sample Output

8
8

Author

UESTC

Source

2016 Multi-University Training Contest 6

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题解：

就是给你 n个数其中（序号i,j）是必选的,(序号k,l)是必不选的.使得其中的子集在n个数中的总权值为m.
例如样例
f(1,2,3,4)=1 f(2,1,3,4)=1 f(1,3,2,4)=1 f(3,1,2,4)=1
f(1,2,4,3)=1 f(2,1,4,3)=1 f(1,3,4,2)=1 f(3,1,4,2)=1 所以答案为8.

个人感想:
看了一下题目,我基本没什么思路,然后喵了一下题解,说是dp,然后我想了一个晚上..我觉得自己真是弱鸡,还是没怎么想到,看了网上题解,可以推出朴素的O(n^3)算法,我尼玛我一点感觉也没,我都不知道怎么推出朴素了..我反而悟出了这道题的前2维的计数,但是我就想不到怎么把必选和不选的排掉.很烦躁.
然后还是看了一下转载了.但是我就想不通最后两维,好艰难啊.
今天早上还是琢磨了一下,我才把它弄懂.

se->select.
dp[i][j][se][nse] 代表 前i个数,和为j,有se个数必选,有nse个必不选的方案数.
是不是很凌乱.我们一步步来,先想前2维.我到不知道为什么别人数是个背包,千万别和背包混一起,我感觉会走弯路,不过确实也类似了.

我的开始是这样想的.
首先 dp[i][j].前i个数和为j,对于当前这个数,如果选择了,就得+dp[i-1][j-a[i]],这时如果不选,+dp[i-1][j].这个先想明白.
然后再想到第3维（第4维也是类似).
dp[i][j][1]+=dp[i-1][j][1] ,首先前[i-1]个数和为j的方案数,而且其中有1个是必选了,但没有选a[i].
dp[i][j][1]+=dp[i-1][j-a[i]][1],首先前[i-1]个数和为j的方案数,而且其中有1个是必选了,,但选了a[i].a[i],不是必选的.
dp[i][j][1]+=dp[i-1][j-a[i]][0],首先前[i-1]个数和为j的方案数,而且其中有1个是必选了,,选了a[i].a[i]是必选的.
这样理解应该懂了吧,其中1维也是这样的思想..这就推出来了几种必选和必不选的方案数了.

分析: 计数dp.

转载：http://blog.csdn.net/zzz805/article/details/52135094

#include <bits/stdc++.h>

using namespace std;

const int mod=1e9+;

int dp[][][][];

int n,s,T;

int a[];

int main()

{

    scanf("%d",&T);

    for(;T>;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++) //前i件物品

            for(int j=;j<=s;j++)  //背包当前容量为j

             for(int x=;x<=;x++) //dp[i][j]x][y]中有x个物品是属于必选的

               for(int y=;y<=;y++) //dp[i][j][x][y]中有y个物品是属于不能选的

        {

            dp[i][j][x][y]=(dp[i][j][x][y]+dp[i-][j][x][y])%mod; //第i件item不选

            if (y->=) dp[i][j][x][y]=(dp[i][j][x][y]+dp[i-][j][x][y-])%mod;

            if (j-a[i]>=)

            {

                dp[i][j][x][y]=(dp[i][j][x][y]+dp[i-][j-a[i]][x][y])%mod;

                if (x->=) dp[i][j][x][y]=(dp[i][j][x][y]+dp[i-][j-a[i]][x-][y])%mod;

            }

        }

      long long ans=;

      for(int j=;j<=s;j++) ans=(ans+dp[n][j][][])%mod;

      ans=(ans*)%mod;

      printf("%lld\n",ans);

    }

    return ;

}

巴特西

hdu 5800 To My Girlfriend(背包变形)

To My Girlfriend

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