Max Sum

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 
线性DP；
#include<iostream> #include <stdio.h>using namespace std;int a[100010];int main(){    int T,N,i,cas=1;    int start,end,temp,sum,max=0;//标记;    scanf("%d",&T);    while(T--){        scanf("%d",&N);        start=end=temp=1;        sum=0;        max=-1001;        for(i=1;i<=N;i++){            scanf("%d",&a[i]);            sum+=a[i];            if(sum>max)            {                max=sum;                start=temp;                end=i;            }            if(sum<0){                sum=0;                temp=i+1;            }        }        printf("Case %d:\n",cas++);        printf("%d %d %d\n",max,start,end);        if(T>0)                    //格式；            printf("\n");    }    return 0;}