Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 294096 Accepted Submission(s): 69830
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
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int num[100086];
int main()
{
int T;
scanf("%d",&T);
int cases=0;
while(T--){
cases++;
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&num[i]);
}
int l=1,r=1;
int x,y;
int maxx=-999999999,sum=0;
for(int i=1;i<=n;i++){
sum+=num[i];
if(sum>maxx){
x=l;y=i;
maxx=sum;
}
if(sum<0){sum=0;l=i+1;}
}
if(cases!=1){printf("\n");}
printf("Case %d:\n",cases);
printf("%d %d %d\n",maxx,x,y);
}
}