2162: Inuyasha And the Monsters

最新推荐文章于 2019-07-23 14:42:00 发布

kongming_acm

最新推荐文章于 2019-07-23 14:42:00 发布

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分类专栏： joj acm_动态规划文章标签： integer kill numbers input ini output

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Inuyasha和Kagome计划清除一系列宫殿中的怪物。这些宫殿从南到北排列成一条直线，他们只能从南向北前进。每个宫殿里藏有一定数量的怪物，如果在当前宫殿杀死的怪物数量超过前方宫殿的数量，则前方的怪物会逃跑。任务是帮助Inuyasha尽可能多地消灭怪物。

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2162: Inuyasha And the Monsters

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	5s	8192K	573	184	Standard

Inuyasha and Kagome, do you hear of them?

Now, they want to kill a great many of monsters. Those monsters hide in several palaces. The palaces are built in a straight line from the south to the north. they must go from the south to the north and can only move straight ahead. That is to say, they can't go from north to the south at any time.

But there is another problem. All the monsters in front of the current palace will escape immedietely if the number of monsters they killed in the current palace is more than the ones of those palaces.

Surely Inuyasha wants to kill as many monsters as he can. Can you help him?

Note: Once they determine to kill the monsters in a palace, they must kill them all.

Input

The input file contains one or more test cases, followed by a line containing the number 0 that signals the end of the file. Each test case consists of two lines.the first line contains a single integer M(M<10000), which indicates the number of the palaces; the second line contains M integers, which are the number of the monsters in the palaces, from south to north. You may assume that all numbers are less than 2^31.

Output

The output for every case begins with a line containing "Case #k:", where k is the number of the test case starting from 1. Then print a single integer which is the number of the monsters they has killed. It is also in a single line. Print a blank line between every two cases.

Sample Input

Sample Output

Case #1:
15

Case #2:
9

Problem Source: chihao

#include<stdio.h>
int dp[10000];
int a[10000];
int main()
{
    int n,i,j;
    int l=1;
    int m=0;
    while(scanf("%d",&n)==1&&n)
    {
        if(m==0) m++;
        else printf("/n");
        printf("Case #%d:/n",l++);
        for(i=0;i<n;i++){scanf("%d",&a[i]);}
        dp[0]=a[0];
        for(i=1;i<n;i++)
        {
            dp[i]=a[i];
            for(j=0;j<i;j++)
            {
                if(a[i]>=a[j]&&dp[i]<dp[j]+a[i]) dp[i]=dp[j]+a[i];
            }
        }
        int max=0;
        for(i=0;i<n;i++)
        {
            if(dp[i]>max) max=dp[i];
        }
        printf("%d/n",max);
    }
    return 0;
}