HDU 1003 Max Sum

题目：

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 -15 4 -7

7 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1: 14 1 4

 

 

Case 2: 7 1 6

题意描述：
输入一串数字为N N(1<=N<=100000)和N个范围-1000到1000的整型数

计算并输出该串数字的最大子串（连续）和以及该子串的起始位置和终止位置

解题思路：
该题还是比较考察思维的，属于简单的DP问题。遍历每个元素，将其累加到sum上，并每次判断sum是否大于max，若大于max则更新sum后，将始末位置也更新一下；若sum<0,表示该数是一个很大的负数需要重新开始累加sum，寻找下一个可能存在的最大子串和。

代码实现：

#include<stdio.h>
int main()
{
    int T,t,n,a[100010],i,sum,s,f,c=0,max;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        sum=0;
        max=-99999999;//注意将max赋值为一个较小的赋值 
        t=1;
        for(i=1;i<=n;i++){
            sum += a[i];
            if(sum > max){
                max=sum;
                s=t;
                f=i;
            }
            if(sum < 0){
                sum=0;
                t=i+1;
            }
        }
        printf("Case %d:
%d %d %d
",++c,max,s,f);
        if(T != 0)
        printf("
");
    }
    return 0;
} 

易错分析：

1、注意格式问题

2、注意max的初始化