Max Sum 最大连续子序列和

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

 
题意很清楚，就是最大子序列和，不过需要输出最大和区间的开始和结束，start为当前起始点

#include <iostream>#include <cstdio>using namespace std;const int N = 1e5 + 5;int a[N];int main(){    int t,n;    cin >> t;    for(int k = 1; k <= t; k++)    {        cin >> n;        for(int i = 1; i <= n; i++)            scanf("%d",&a[i]);        int maxn = -1e8,sum = 0;    //maxn为当前最大值        int p,q,start = 1;           //起始点一开始为1，p为起始，q为结束        for(int i = 1; i <= n; i++)        {            sum += a[i];            if(sum > maxn)            {                maxn = sum;                p = start;                q = i;            }            if(sum < 0)            {                sum = 0;                start = i+1;         //更改起始点            }        }        printf("Case %d:\n",k);        printf("%d %d %d\n",maxn,p,q);        if(k < t)            printf("\n");    }    return 0;}