J - Max Sum

J - Max Sum

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

#include <iostream>#include <cstdio>#define  MAX 100005using namespace std;int a[MAX];int main(){int T,N,t=0,sum[MAX],s[MAX],ans,tt;cin>>T;tt=T;while(tt--){cin>>N;for(int i=0;i<N;i++)cin>>a[i];sum[0]=a[0];s[0]=0;ans=0;for(int i=1;i<N;i++){if(sum[i-1]>=0){sum[i]=sum[i-1]+a[i];s[i]=s[i-1];}else{sum[i]=a[i];s[i]=i;} if(sum[ans]<sum[i]) ans=i;}printf("Case %d:\n%d %d %d\n",++t,sum[ans],s[ans]+1,ans+1);if(t!=T)cout<<endl;}return 0;}