PAT 1049. 数列的片段和(20)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;


int main() {
	int n;
	double num, sum = 0;
	cin >> n;
	for ( int i = 1; i <= n; i++ ) {
		cin >> num;
		sum += num * i * (n - i + 1);
	}
	printf("%.2lf\n", sum);
	return 0;
}