hdu1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 67750    Accepted Submission(s): 15546

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

 

这是一个最最大子段和的问题。并且输出子段的头根尾部。

#include<iostream>
#include<stdio.h>
using namespace std;
int a[100000+10];
int sum[100000+10];
void maxsum(int n)
{
    sum[0]=-1;
    sum[1]=a[1];
    int max=a[1];
    int right=1;
    for(int i=2;i<=n;i++)
    {
        if(sum[i-1]>0)sum[i]=sum[i-1]+a[i];
        else sum[i]=a[i];
        if(max<sum[i])
        {
            max=sum[i];right=i;
        }
    }
    int left=right;
    while(sum[left]>=0)left--;
    if(left!=right)//注意这里要不然rihgt就会大于left了。
    left++;
    printf("%d %d %d\n",max,left,right);
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        int n;
        scanf("%d",&n);
        for(int j=1;j<=n;j++)
        {
            scanf("%d",&a[j]);
        }
        printf("Case %d:\n",i);
        maxsum(n);
        if(i!=t)printf("\n");
    }
    return 0;
}