hdu 1003 Max Sum

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本文解析了一个经典的编程问题——最大子序列和问题，通过示例详细介绍了如何使用前缀和的方法来高效求解该问题，并提供了完整的代码实现。

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 110252 Accepted Submission(s): 25443

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

我们用s[i] 表示a[i]的前i项和 ,s[i]=sum(a[k])(k=1,2,.....i)

然后遍历一遍s[],记录最小的s[i]值，存于min,那么最大子段和就是 max=Max(s[i]-min);

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int a[100100],sum[100100];
int l,r,mins,minp,ans;

int main()
{
    int t,i,k,n;
    scanf("%d",&t);
    for(k=1; k<=t; k++)
    {
        scanf("%d",&n);
        memset(sum,0,sizeof sum);
        for(i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        l=r=minp=mins=0;
        ans=0xcfcfcfcf;
        for(i=1; i<=n; i++)
        {
            if(sum[i]-mins>ans)
            {
                ans=sum[i]-mins;
                l=minp+1;
                r=i;
            }
            if(sum[i]<mins)
            {
                mins=sum[i];
                minp=i;
            }
        }
        printf("Case %d:\n",k);
        printf("%d %d %d\n",ans,l,r);
        if(k!=t)
            printf("\n");
    }
    return 0;
}