HDU 1003 Max Sum

最新推荐文章于 2022-03-03 08:27:19 发布

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最新推荐文章于 2022-03-03 08:27:19 发布

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分类专栏： dp 文章标签： dp

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Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 137183 Accepted Submission(s): 31788

Problem Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input

  
  
   
   2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output

  
  
   
   Case 1:
14 1 4

Case 2:
7 1 6

跟上一题一样。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int a[100005];
int dp[100005];

int main()
{
    int t,i,j,n;
    scanf("%d",&t);

    for(i=1;i<=t;i++)
    {
        scanf("%d",&n);
        for(j=1;j<=n;j++)
            scanf("%d",&a[j]);

        dp[1]=a[1];
        int r=1,l=1,current=1;
        int ma=-1;

        for(j=2;j<=n;j++)
        {
            if(dp[j-1]>=0)//确定dp[j]的值
                dp[j]=dp[j-1]+a[j];

            else
            {
                dp[j]=a[j];
                current=j;//current是个中间变量,保存起点的值
            }//先求出dp[j]的大小然后跟ma比较，在判断起点终点，否则是没有意义的，所以先用current保存起点的值，而终点的值就是当前值，更新的时候起点不一定换但终点一定换
            if(dp[j]>=ma)//确定起点终点，这个=要有，没有会wa
            {
                ma=dp[j];
                r=j;
                l=current;
            }
        }

        printf("Case %d:\n",i);
        printf("%d %d %d\n",ma,l,r);
        if(i!=t)
        printf("\n");
    }
    return 0;
}