杭电 1003

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
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6

代码：

#include<stdio.h>int main(){    int t,max=0,i,n,m,r,l,sum=0,k=1;    scanf("%d",&t);    int s=1;    while(t--)    {        sum=0;max=-1001;l=r=1;k=1;        scanf("%d",&n);        for(i=1;i<=n;i++)        {scanf("%d",&m);        sum+=m;        if(sum>max)        {max=sum;r=i;l=k;}        if(sum<0)        {sum=0;k=i+1;}        }           printf("Case %d:\n",s);        s++;        if(t!=0)        printf("%d %d %d\n\n",max,l,r);        else                printf("%d %d %d\n",max,l,r);    }    return 0;}