Given a sequence of K integers { N​1​​, N​2​​, ..., N​K​​ }. A continuous subsequence is defined to be { N​i​​, N​i+1​​, ..., N​j​​ } where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.（翻译：如果所有K个数都是负数，那么它的最大和被定义为0，你就应该输出整个序列的第一个和最后一个数。hh）

Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

1. //输入数列个数，和数列内容
2. //输出最大子列和，以及这个最大子列的第一个和最后一个数
3. //MaxSum<0即任意序列和均小于0，则K个整数均为负数，此时最大子列和为0，第一和最后一个就是数列的首位

#include <stdio.h>void MaxSubSum(int a[],int n);int main(){int n;scanf("%d",&n);//输入数列的个数int a[n];for(int i=0;i<n;i++){scanf("%d",&a[i]);//输入数列 } MaxSubSum(a,n);return 0;
} void MaxSubSum(int a[],int n){int f=0,temp_f=0,l=0;
//first和last，临时first，因为如果后面出现了更大的子列和就需要修改int thisSum=0,maxSum=-1; 
//maxSum初始化为零的话，当全部是负数或者0时if(maxSum<0)这个判断条件永远不会被执行，出现死语句。for(int i=0;i<n;i++){thisSum+=a[i];//向右累加if(thisSum<0){thisSum=0;//负值对最大子列和没有帮助，抛弃从0开始往后累加temp_f=i+1;	//暂且认为下一个数会是最大子列的第一个数		}else if(thisSum>maxSum){maxSum=thisSum;f=temp_f;//如果thisSum>maxSum,则temp_f确实是最大子列的第一个数l=i;//当前的位置就是最后一个数的下标									}	}if(maxSum<0){printf("0 %d %d",a[0],a[n-1]) ;
//不能因为题目是因为的就只看输入输出啊！题目要求如果全为负数，那么以这种格式输出}else{printf("%d %d %d",maxSum,a[f],a[l]) ;}}