PAT甲级A1104 Sum of Number Segments （20 分）

本文链接：https://blog.youkuaiyun.com/vir_lee/article/details/87937058

本文介绍了一种计算给定正数序列所有子序列和的方法。通过分析序列中每个数字出现的次数和位置，得出一个公式来快速计算总和，避免了暴力求解的复杂度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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) and (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

题意：给出n个数，求这n个数所有子序列的和

思路：由题可得规律，第一个数字出现一轮，共4次，第二个数字出现两轮，每轮3次，第三个数字出现三轮，每轮2次，第四个数字出现四轮，每轮1次。

参考代码：

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