HDOJ 5000 Clone

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 所有滿足的情況的屬性和是一定的,而且屬性和等於sum/2時得到的結果最大.

CloneTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 574    Accepted Submission(s): 277


Problem DescriptionAfter 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. 
InputThe 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]. 
 
OutputFor each test case, output an integer representing the answer MOD 10^9 + 7. 
Sample Input

21528 6

 
Sample Output

17

 
Source2014 ACM/ICPC Asia Regional Anshan Online 

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long int LL;const LL mod=(1e9+7);int v[2020],n;LL sum,dp[2][2000400];int main(){    int T_T;    scanf("%d",&T_T);    while(T_T--)    {        scanf("%d",&n);        sum=0;        for(int i=0;i<n;i++)        {            scanf("%d",v+i);            sum+=v[i];        }        memset(dp,0,sizeof(dp));        for(int i=0;i<n;i++)        {            if(i==0)            {                for(int j=0;j<=v[i];j++) dp[0][j]=1;                continue;            }            for(int j=0;j<=sum/2;j++)            {                int temp=0;                for(int k=0;k<=v[i]&&k<=j;k++)                {                    temp=(temp+dp[(i%2)^1][j-k])%mod;                }                dp[i%2][j]=temp;            }        }        printf("%d\n",dp[(n-1)%2][sum/2]);    }    return 0;}

HDOJ 5000 Clone