HDU 1003（A - 最大子段和）

HDU   1003（A - 最大子段和）

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87125#problem/A

 

题目：

          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

 

 

题意：

给出一个序列，求此序列的最大子段和 及开始和结束的位置。

 

分析：

动态规划。求最大子段和。b[i]表示最大子段和。c[i]表示子段和开始的位置。b[i]=(b[i-1]+a[i])>a[i]?b[i-1]+a[i]:a[i]

 

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int maxn=100005;

int a[maxn],b[maxn],c[maxn];//b表示最大子段和，c表示开始的位置

int main()
{
    int t,m=1;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        b[1]=a[1];//记开始时的子段和为b[1]
        c[1]=1;//开始时的位置为1
        for(int i=2;i<=n;i++)
        {
            if(b[i-1]>=0)
            {
                b[i]=b[i-1]+a[i];//子段和
                c[i]=c[i-1];
            }
            else
            {
                b[i]=a[i];
                c[i]=i;
            }
        }
        int max=b[1];//令起始位置最大值为b[1]
        int end=1;
        for(int i=2;i<=n;i++)
        {
            if(b[i]>max)
            {
                max=b[i];
                end=i;//结束位置
            }
        }
        printf("Case %d:\n",m++);
        printf("%d %d %d\n",max,c[end],end);
        if(t)
            printf("\n");
    }
    return 0;
}

 

我先在杭电上做了几遍，提交一直都是WA，改了几次都是WA。第一次是因为我直接把开始位置写为1，根本没有计算c[i]。后来又出现了PE，因为我没有写if(t)。改了几次终于改成功了。