HDU1003--Max Sum--最大连续和

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>using namespace std;int A[100008];#define INF 0x3fint maxmax(int a,int b){return a>b?a:b;}int main(){int t;scanf("%d",&t);for(int i=1;i<=t;i++){int n;scanf("%d",&n);for(int j=1;j<=n;j++){scanf("%d",&A[j]);}int sum=0,max=-INF;int head=1,tail=1;for(int j=1;j<=n;j++){sum+=A[j];if(max<sum){tail=j;max=sum;}sum=sum<0?0:sum;}int sum1=0;for(int j=1;j<=n;j++){sum1+=A[j];if(sum1==max){break;}if(sum1<0){sum1=0;head=j+1;}}if(i!=t)cout<<"Case "<<i<<":"<<endl<<max<<" "<<head<<" "<<tail<<endl<<endl;else cout<<"Case "<<i<<":"<<endl<<max<<" "<<head<<" "<<tail<<endl;}return 0;}