PAT 乙级 1049 数列的片段和 (20分)

#include <iostream>
using namespace std;

int main() {
	int n;
	cin >> n;
	double sum = 0.0, temp;
	for (int i = 1; i <= n; i++) 
	{
		cin >> temp;
		sum = sum + temp * i * (n - i + 1);
	}
	printf("%.2f", sum);
	system("pause");
	return 0;
}

标签：%.,PAT,temp,int,1049,cin,20,sum
来源： https://blog.csdn.net/qq_43180487/article/details/110243975