Max Sum 最大连续和的子序列

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：给n个数字，求出n个数字最大的连续和，起始位置和终点位置是多少。
思路：DP，当数字和小于0的时候重置，并记录起始位置，如果大于ans，则更新终止位置
#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>using namespace std;int main(){    int n,t,icase=1;    scanf("%d",&t);    for(int icase=1;icase<=t;icase++)    {        scanf("%d",&n);        int ans=-100000000,s=-100000000,elem,st,en,mid;        for(int i=1;i<=n;i++){            scanf("%d",&elem);           if(s<0){            s=elem;            mid=i;           }            else             s+=elem;            if(s>ans) ans=s,en=i,st=mid;        }        printf("Case %d:\n",icase);        printf("%d %d %d\n",ans,st,en);        if(icase!=t)        printf("\n");    }    return 0;}