C++ Pat甲级1007 Maximum Subsequence Sum （25 分）(dp)

1007 Maximum Subsequence Sum （25 分）

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 Knumbers, 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.

Sample Input:

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

Sample Output:

10 1 4

输入：第一行数字总数k，

  第二行 k个数

输出：这k个数的最大子序列和，以及此序列的起始值和末尾值的

转：

分析：sum为要求的最大和，temp为临时最大和，leftindex和rightindex为所求的子序列的下标，tempindex标记left的临时下标～
temp = temp + v[i]，当temp比sum大，就更新sum的值、leftindex和rightindex的值；当temp < 0，那么后面不管来什么值，都应该舍弃temp < 0前面的内容，因为负数对于总和只可能拉低总和，不可能增加总和，还不如舍弃～
舍弃后，直接令temp = 0，并且同时更新left的临时值tempindex。因为对于所有的值都为负数的情况要输出0，第一个值，最后一个值，所以在输入的时候用flag判断是不是所有的数字都是小于0的，如果是，要在输入的时候特殊处理～

 
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作者：柳婼 
来源：CSDN 
原文：https://blog.csdn.net/liuchuo/article/details/52144554 
版权声明：本文为博主原创文章，转载请附上博文链接！

#include <iostream>
#include <vector>
using namespace std;
int main() {
	int n;
	scanf("%d", &n);
	vector<int> v(n);
	int leftindex = 0, rightindex = n - 1, sum = -1, temp = 0, tempindex = 0;
//动态规划 
	for (int i = 0; i < n; i++) {
		scanf("%d", &v[i]);
		temp = temp + v[i]; //临时和 
		if (temp < 0) { //小于0则从下一个数重新开始计算 
			temp = 0;  
			tempindex = i + 1;
		} else if (temp > sum) { //临时和大于当前的sum，则更新sum ,最左边的值的位置为上一次tempindex,最右边的值的位置为当前的i 
			sum = temp;
			leftindex = tempindex;
			rightindex = i;
		}
	}
	if (sum < 0) sum = 0;
	printf("%d %d %d", sum, v[leftindex], v[rightindex]);
	return 0;
}