1104. Sum of Number Segments (20)

#include<bits/stdc++.h>
using namespace std;
double n[100003];
//int time[100003];
intmain()
{
	int N;
	scanf("%d",&N);	
	for(int i=1;i<=N;i++)
		scanf("%lf",&n[i]);
	double sum=0;
	for(int i=1;i<=N;i++){
		sum+=n[i]*i*(N-i+1);//***
	}
	sum=(sum*100+0.5)/100;
	printf("%.2f\n",sum);
	return 0;
}

The position of the asterisk in the code needs to be noted that if the integer is multiplied first and then the floating-point number is multiplied, then the amount of data of 1e5 is easy to overflow and cause an error.

Converting to floating point numbers is fine

I've been looking for a long time and I don't know what's wrong....-- I thought it was a precision pot