hud1003最大子序列和

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52833    Accepted Submission(s): 11726

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
/* 
最大连续子序列  动态规划 
 
<a target=_blank href="http://acm.hdu.edu.cn/showproblem.php?pid=1231">http://acm.hdu.edu.cn/showproblem.php?pid=1231</a> 
 
题意：给你一系列整数，选出所有 连续子序列 中 元素和最大 的一个子序列， 
输出最大和、最大连续子序列的第一个和最后一个元素 
 
思路见注释 
*/  
  
#include <algorithm>  
#include <iostream>  
#include <iomanip>  
#include <fstream>  
#include <cstring>  
#include <string>  
#include <cstdio>  
#include <cmath>  
#include <stack>  
  
using namespace std;  
const int MAX=10005;  
  
int a[MAX];  
  
int main()  
{  
    int n;  
    int flag,max,sum,start,end,temp;  
    int i,j;  
    while(cin>>n && n){  
         flag=0;  
         for(i=0;i<n;i++)  
         {  
             scanf("%d",&a[i]);  
             if(a[i]>=0)  
                 flag=1;  
         }  
         max=0;  
         if(flag==0)  
         {  
             printf("%d %d %d\n",max,a[0],a[n-1]);  
             continue;  
         }  
         else  
         {  
            sum=a[0];  
            max=a[0];  
            start=0;  
            end=0;  
            temp=0;  //临时起点  
            for(i=1;i<n;++i)  
            {  
                if(sum<0) { //如果前几项和小于0，就从新开始记录  
                    sum=0;  
                    temp=i; //刷新临时起点  
                }  
                sum+=a[i];                
                if(sum>max){ //如果现在的sum大于原来的max,刷新数据  
                    max=sum;  
                    start=temp;  
                    end=i;  
                }  
            }  
            printf("%d %d %d\n",max,a[start],a[end]);  
         }  
    }  
}