【PAT】【Advanced Level】1104. Sum of Number Segments (20)

最新推荐文章于 2024-05-11 08:07:41 发布

原创最新推荐文章于 2024-05-11 08:07:41 发布 · 374 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

132 篇文章

订阅专栏

本文介绍了一种针对正数序列求所有连续子序列（段）之和的算法，并提供了具体的实现代码。通过分析每个元素在不同子序列中出现的次数规律，实现了高效的求和计算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1104. Sum of Number Segments (20)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

Given a sequence of positive numbers, a segment is defined to be a consecutive subsequence. For example, given the sequence {0.1, 0.2, 0.3, 0.4}, we have 10 segments: (0.1) (0.1, 0.2) (0.1, 0.2, 0.3) (0.1, 0.2, 0.3, 0.4) (0.2) (0.2, 0.3) (0.2, 0.3, 0.4) (0.3) (0.3, 0.4) (0.4).

Now given a sequence, you are supposed to find the sum of all the numbers in all the segments. For the previous example, the sum of all the 10 segments is 0.1 + 0.3 + 0.6 + 1.0 + 0.2 + 0.5 + 0.9 + 0.3 + 0.7 + 0.4 = 5.0.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N, the size of the sequence which is no more than 10⁵. The next line contains N positive numbers in the sequence, each no more than 1.0, separated by a space.

Output Specification:

For each test case, print in one line the sum of all the numbers in all the segments, accurate up to 2 decimal places.

Sample Input:

4
0.1 0.2 0.3 0.4

Sample Output:

5.00

原题链接：

https://www.patest.cn/contests/pat-a-practise/1104

思路：

找规律

对于一个长度为n的序列

第1个元素，累加了n*1次；

第2个元素，累加了（n-1）*2次；

。。。

第n个元素，累加了1*n次。

按此规律，边读入边计算即可

CODE：

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
	int n;
	double res=0;
	scanf("%d",&n);
	for (int i=0;i<n;i++)
	{
		double t;
		scanf("%lf",&t);
		res+=(t*(double)(n-i)*(double)(i+1));
	}
	printf("%.2f",res);
	return 0;
}