PAT 1049.数列的片段和

求所有连续字段的和，网上说实际上求每个数出现的次数

每个数的出现次数为（n-i）*(i+1)

#include<stdio.h>#include<iostream>using namespace std; int main(){int n,i;double sum = 0,a[100000];cin >> n;for(i=0;i<n;i++){cin >> a[i];}for(i=0;i<n;i++){sum += (double)(n-i)*(i+1)*a[i];}printf("%.2lf\n",sum);}