A

A - Max Sum

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

Submit 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

 25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5 

 

Sample Output

 Case 1:14 1 4Case 2:7 1 6 

 

此题为最大子列和并输出此子列和首尾坐标

#include<iostream>#include <algorithm>#include<string.h>#include<stdio.h>using namespace std;int main(){int n,h=1;cin>>n;while(n--){int m,sum,i,j;cin>>m;int a[100005],b[100005]={0};int c=-0xffffff;for(i=1;i<=m;i++){cin>>a[i];}for(i=1;i<=m;i++){if(b[i-1]+a[i]>a[i])b[i]=b[i-1]+a[i];elseb[i]=a[i];if(c<b[i]){c=b[i];sum=i;}}for(i=sum;i>0;i--){if(b[i-1]!=0&&b[i]==a[i])break;}if(i==0)i++;printf("Case %d:\n",h++);cout<<c<<" "<<i<<" "<<sum<<endl;if(n!=0) cout<<endl;}}