hdu 5000 Clone 2014 ACM/ICPC Asia Regional Anshan Online

Clone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 447    Accepted Submission(s): 215

Problem Description

After eating food from Chernobyl, DRD got a super power: he could clone himself right now! He used this power for several times. He found out that this power was not as perfect as he wanted. For example, some of the cloned objects were tall, while some were short; some of them were fat, and some were thin. 

More evidence showed that for two clones A and B, if A was no worse than B in all fields, then B could not survive. More specifically, DRD used a vector v to represent each of his clones. The vector v has n dimensions, representing a clone having N abilities. For the i-th dimension, v[i] is an integer between 0 and T[i], where 0 is the worst and T[i] is the best. For two clones A and B, whose corresponding vectors were p and q, if for 1 <= i <= N, p[i] >= q[i], then B could not survive. 

Now, as DRD's friend, ATM wants to know how many clones can survive at most.

 

Input

The first line contains an integer T, denoting the number of the test cases.

For each test case: The first line contains 1 integer N, 1 <= N <= 2000. The second line contains N integers indicating T[1], T[2], ..., T[N]. It guarantees that the sum of T[i] in each test case is no more than 2000 and 1 <= T[i]. 

 

Output

For each test case, output an integer representing the answer MOD 10^9 + 7.

 

Sample Input

2 1 5 2 8 6

 

Sample Output

1 7

 

Source

2014 ACM/ICPC Asia Regional Anshan Online

 

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題意：一個克隆人有n個屬性，如果一個克隆人的任何一個屬性都不大於另一個，他就不應該存在，給你n和每個屬性的最大值，問最多能有多少克隆人存在。

思路：如果說能保證最多的克隆人存在，那麼他們的屬性和一定相等，且爲所有最大屬性和的一半。（好像很有道理，卻說不出來哪有道理的結論）。比賽的時候神犇推得出來揹包一下就過掉了，我這樣的小菜做這種題只能靠yy了。知道結論直接01揹包求能到達sum/2的所有情況就好了。

#include <iostream>
#include <stdio.h>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int mod=1e9+7;
const int N=2005;
int dp[N*N/2],a[N];
int main()
{
    int n,t,sum;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        sum=0;
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        sum/=2;
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=0;i<n;i++){
            for(int j=sum;j>=0;j--){
                for(int k=1;k<=a[i]&&k<=j;k++)
                 dp[j]=(1ll*dp[j]+dp[j-k])%mod;
            }
        }
        printf("%d\n",dp[sum]);
    }
    return 0;
}