Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2121 Accepted Submission(s): 798 

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

题目大意：最大字段和问题   加了一个下标  即输出 这个字段和的其实位置
注意看样例中csae 2  最大和是7  是从1-6 也就是说前面存在一段和等于0 也加上  具体看代码
#include<iostream>#include<string.h>#include<stdio.h>using namespace std;int main(){    int a[100001],st,end;    int n,i,j,s,max,sum;    int t,tt;    int cc=1;    cin>>t;    int ttt=t;    while(t--)    {        cin>>n;        st=1;        tt=1;        sum=0;        memset(a,0,sizeof(a));        max=-1000000;        for(i=0;i<n;i++)        {           scanf("%d",&a[i]);            sum+=a[i];            if(sum>max)            {                max=sum;                st=tt;                end=i+1;//i从0 开始的            }            if(sum<0)  // 舍弃 更新当前开始位置+1            {                 tt=i+2;                sum=0;            }        }        cout<<"Case "<<cc<<":"<<endl;        cout<<max<<" "<<st<<" "<<end<<endl;        cc++;        if(t)cout<<endl;//输出陷阱    }    return 0;}