HDU 1003 Max Sum(连续子列的最大和，动态规划）

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55792    Accepted Submission(s): 12580

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

 

Author

Ignatius.L

算法：1.一直相加直到和出现负数，在相加期间能得到一个最大和，记录
2.重复1步骤，更新最大和，直到输入完毕
原理：如果在连续子串 <a1,a2,a3,...,an> 中Sn<0;S1,S2,...S(n-1)均大于或等于0
那么肯定存在一个最大的Si,Si也就是a1到an这段连续子串的最大和。
我们知道a1>0时任意的 <ai...an> (i>1) 的和都小于0，
所以这些元素就不能与an后的元素构成能获得最大和的子串(如果a1=0则a2>=0,情况类似);
同时在 <ai...aj> (i>1,j<=n) 也不能或得比Si大的和。

/*
HDU1003
动态规划
*/ 
#include<stdio.h>
#define MAXN 100000
int main()
{
    int T,iCase,i,sum,maxsum,j,x,y,a;
    int n;
    scanf("%d",&T);
    iCase=0;
    while(T--)
    {
        iCase++;
        sum=0;
        maxsum=-20000;
        i=1;
        scanf("%d",&n);
        for(j=1;j<=n;j++)
        {
            scanf("%d",&a);
            sum+=a;
            if(maxsum<sum)
            {
                maxsum=sum;
                x=i;
                y=j;
            }    
            if(sum<0)
            {
                i=j+1;
                sum=0;
            }    
        }
        printf("Case %d:\n",iCase);
        printf("%d %d %d\n",maxsum,x,y);    
        if(T>0)printf("\n");
    }    
    return 0;
}