H题 Max sum

Problem H: 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

题意简单，找到一串数字中的最长子串 。困扰我的主要就是就是几组数据 1 2 -3 ,3和 0 0 -4 0 结果跟我预判的大不一样：分别是：3 1  2；0  1 1；还有两种没考虑到的情况：-3和-4，-3，-2，-1；结果是：-3  1 1；-1 4 4

结果就被坑了；另一个问题，就是第二个if ，不能用else if ，排除全是负数的情况。

算法就是在读入数据时记录临时最大值，并保持更新；

#include <stdio.h>#include <stdlib.h>int main(){    int t,T;    scanf("%d",&t);    T=t;    while(T--)    {        int n,i,maxn=-1001,tempbeg=0,temp=0,beg=0,en=0;//maxn是最大值，beg为起点，en为终点；tempbeg是临时最大值起点；千万别忘了要复制给maxn一个最小值！        scanf("%d",&n);        int a[n];        for(i=0;i<n;i++)            scanf("%d",&a[i]);        for(i=0;i<n;i++)        {            temp+=a[i];            if(temp>maxn)               {                   en=i;                   beg=tempbeg;                   maxn=temp;               }            if(temp<0)            {                    tempbeg=i+1;                  temp=0;            }        }         printf("Case %d:\n%d %d %d\n",t-T,maxn,beg+1,en+1);         if(T!=0)          putchar('\n');    }    return 0;}