[HDU1003]Max Sum

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

My Problem Report

这是一道经典的基础动态规划题目，因此网上大多数的题解都是DP，只需要推导出简单的动归方程f[i]=Max{f[i-1]+a[i],a[i]}即可。
因此我在这里只详细介绍 模拟方法：

此算法效率与DP相同，为是O(n)，代码复杂度也不高。算法的核心思想是：最大字串＝Max｛左端字串－左端字串的左端最小子串｝
我们只需从左到右扫描一次，每一次处理首先纪录前i项和
重点在第二步，我们不能先先求左端最小子串，否则会导致当输入数据全为负的输出结果为错误答案零
我们应到首先进行MAX操作，并纪录起始结束点，最后在更新左端最小子串
下面分别附上我的模拟代码和DP代码，仅供大家参考。

Code I

//  Created by Chlerry on 14-10-22.
//  Copyright (c) 2014 Chlerry. All rights reserved.
//  http://acm.hdu.edu.cn/showproblem.php?pid=1003

#include 
#include 
#include 
#include 

int t,n;
int a[100010];

int main(int argc,const char *argv[])
{
    //freopen("out.txt","w",stdout);
    scanf("%d",&t);
    for(int ca=1;ca<=t;ca++)
    {
        int min=INT_MAX,minn;
        int max=INT_MIN,start,end;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]+=a[i-1];
            if(min>=0 && max1;
                end=i;
            }
            if(min<0 && max1;
                end=i;
            }
            if(min>a[i])
            {
                min=a[i];
                minn=i;
            }
        }
        printf("Case %d:\n%d %d %d\n",ca,max,start,end);
        if(ca!=t)
            printf("\n");
    }
    return 0;
}

Code II

//  Created by Chlerry on 14-10-28.
//  Copyright (c) 2014 Chlerry. All rights reserved.
//  http://acm.hdu.edu.cn/showproblem.php?pid=1003

#include 
#include 
#include 

int a[100010],f[100010],start[100010];

int main(int agrc,const char *argv[])
{
    int t,n;
    int max,maxn;
    scanf("%d",&t);
    start[0]=1;
    for(int ca=1;ca<=t;ca++)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(f[i-1]>0)
            {
                f[i]=f[i-1]+a[i];
                start[i]=start[i-1];
            }
            else
            {
                f[i]=a[i];
                start[i]=i;
            }
        }
        max=INT_MIN;
        for(int i=1;i<=n;i++)
            if(maxprintf("Case %d:\n%d %d %d\n",ca,max,start[maxn],maxn);
        if(ca!=t)
            printf("\n");
    }   
    return 0;
}