HDU-1003 Max Sum

A - Max Sum 
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u 
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Status

Practice

HDU 1003 
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

#include<stdio.h>int main(){int a[100100],t,n,sum,f,ans,k,z,w=0,c;scanf("%d",&t);c=t;while(t--){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);sum=0;k=f=z=1;ans=a[1];for(int i=1;i<=n;i++){if(sum>=0)sum+=a[i];else{sum=a[i];k=i;}if(ans<sum){ans=sum;z=i;f=k;}}printf("Case %d:\n%d %d %d\n",++w,ans,f,z);if(w<c) printf("\n");}}