wordpress模板layuiseo应该如何做
强烈推荐,刷PTA的朋友都认识一下柳神–PTA解法大佬
本文由参考于柳神博客写成
柳神的CSDN博客,这个可以搜索文章
柳神的个人博客,这个没有广告,但是不能搜索
还有就是非常非常有用的 算法笔记 全名是
算法笔记 上级训练实战指南 //这本都是PTA的题解
算法笔记
PS 今天也要加油鸭
题目原文
Given a sequence of K integers { N1, N2, …, 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.
Sample Input:
10
-10 1 2 3 4 -5 -23 3 7 -21
Sample Output:
10 1 4
生词如下:
PS:基本看懂了.因为原来MOOC上有一题,一模一样,但是我写的还是不好,
直接用柳神的了.
smallest indices i and j 最小索引的 i 和 j
思路如下: Copy 柳神
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～
代码如下:
#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;for (int i = 0; i < n; i++) {scanf("%d", &v[i]);temp = temp + v[i];if (temp < 0) {temp = 0;tempindex = i + 1;}else if (temp > sum) {sum = temp;leftindex = tempindex;rightindex = i;}}if (sum < 0) sum = 0;printf("%d %d %d", sum, v[leftindex], v[rightindex]);return 0;
}
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