PAT A1104 2019.07.16 【水题】

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本文介绍了一种针对数列的特殊求和方法——求所有连续子序列（段）元素之和的问题解决思路及实现代码。通过分析序列中每个元素在不同子序列中的贡献度，实现了高效的求和算法。

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1104 Sum of Number Segments (20 分)

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

思路分析

其实就是初中做的找规律题，无非是变成用代码来实现

我是直接加起来的，每次循环都少了n个ni

但是只对了3个点，思路应该没有问题，就这样吧

#include<cstdio>
#include<iostream>
#define MAX 100010
double arr[MAX]={0.0};
double temp[MAX]={0.0};

int main()
{
	int n;
	scanf("%d",&n);
	if(n<=0)return 0;
	
	double sum=0;				 
	for(int i=0;i<n;i++)
	{
		scanf("%lf",&arr[i]);
		if(i==0)temp[i]=arr[i];										
		else temp[i]=temp[i-1]+arr[i]; 
		sum=sum+temp[i];
	}	
	
	long double output=0;
	double sub=0;
	for(int i=0;i<n;i++)
	{
		output=output+sum-sub;
		sub=sub+(n-i)*arr[i];
	}
	
	
	printf("%0.2llf",output);														 
	
	
	
}