1002 Max Sum

Max Sum

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

 

Author

Ignatius.L

 

先顺序求最大值 找到最大的和 标记末值。再反过来从末值减回去到sum变回0，标记初值。

#include <stdio.h>int main(){    int t;    scanf("%d",&t);    for(int v=1;v<=t;v++)    {        if(v!=1)printf("\n");        int n,a[100100]={0};        scanf("%d",&n);        int sum=0,max=-2000,begin=1,end=1;        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            sum=sum+a[i];            if(sum>max)max=sum,end=i;            if(sum<0)sum=0;        }//求出最大值和末值        sum=max;        for(int i=end;i>0;i--)        {            sum=sum-a[i];            if(sum==0)begin=i;        }        printf("Case %d:\n",v);        printf("%d %d %d\n",max,begin,end);    }    return 0;}