5thweek.problem_A hdu1003最大子段和

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

 

分析：

DP题。状态转移方程：dp[i]=d[i-1]+a[i]>a[i]?d[i-1]+a[i]:a[i];

 

错误代码如下：（关键错误和正确代码的对比）

#include<iostream>
#include<cstdio>
using namespace std;
const int m=100005;
int a[m];

int main()
{
    int t,k=0;
    scanf("%d",&t);
    while(t--)
    {
        int n,i,maxn,before,begin=1,end=1;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {

            scanf("%d",&a[i]);
            if(i==1)
             {
                 maxn=a[i]; before=a[i];
             }
             else
             {
                if(before+a[i]<a[i])
                {
                     before=a[i];
                     begin=i;
                }
                else
                {
                   before+=a[i];
                }

             }
             if(before>maxn)
             {
                  maxn=before;
                  //x=begin;
                  end=i;
             }

        }
        k++;
        printf("Case %d:
",k);
        printf("%d %d %d
",maxn,begin,end);
        if(t)
            printf("
");

    }

    return 0;
}

正确代码如下：（用一个特殊点的案例6      :2 3 -6 4 3 -8找出错误！加入一个关键变量j记录当前a[i]位置）

#include<iostream>
#include<cstdio>
using namespace std;
const int m=100005;
int a[m];

int main()
{
    int t,k=0;
    scanf("%d",&t);
    while(t--)
    {
        int n,i,maxn,before,begin=1,end=1,j=1;  //必须用一个j记录当前数a[i]的位置
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {

            scanf("%d",&a[i]);
            if(i==1)
             {
                 maxn=a[i]; before=a[i];

             }
             else
             {
                if(before+a[i]<a[i])
                {
                     before=a[i];
                     j=i;     //记录当前位置，此时begin不改变
                }
                else
                {
                   before+=a[i];
                }

             }
             if(before>maxn)
             {
                  maxn=before;
                    begin=j;  //只有当before>maxn时再改变
                  end=i;
             }

        }
        k++;
        printf("Case %d:
",k);
        printf("%d %d %d
",maxn,begin,end);
        if(t)
            printf("
");

    }

    return 0;
}