Max Sum||HDU1003

link：http://acm.hdu.edu.cn/showproblem.php?pid=1003
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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

题解：最大子序列和
这道题需要记录这段子序列的起始和结束位置，注意需要在判断前给位置变量赋初值1，不然会WA
AC代码：

#include<cstdio>#include<cstdlib>#include<iostream>#include<algorithm>#include<cmath>#include<cstring>using namespace std;int num[100010];int main(){    int t,c=1,n,x1,y2,x2;    scanf("%d",&t);     while(t--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&num[i]);        }        long long sum=0;        long long Max=-0x3f3f3f3f;        x2=y2=x1=1;//赋初值，x1是重新开一个字段时存储起始位置的         for (int i=1,x1=1;i<=n;i++)        {             sum +=num[i];              if(sum>Max)//当前字段大于已有的最大和时             {                  Max=sum;                  x2=x1;//记录起始位置                 y2=i;//记录结束位置             }              if(sum<0)//小于0尝试重新开一个字段             {                  sum=0;                  x1=i+1;//起始位置从下一个元素开始             }          }        printf ("Case %d:\n%lld %d %d\n",c++,Max,x2,y2);            if(t) printf("\n");    }    return 0;}