1104 Sum of Number Segments (20 point(s))

1104 Sum of Number Segments (20 point(s))

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 10​5​​. 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

数学题。 

#include <stdio.h>
int n;
double x;
int main()
{
    double ans;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lf",&x);
        ans+=x*i*(n-i+1);
    }
    printf("%.2f\n",ans);
    return 0;
}

暴力求解（15分，case#2和3出现TLE）

#include<iostream>
using namespace std;
const int MAX = 1e5+13;
double a[MAX]={0};
double sum[MAX]={0};
int main(void){
	int n;cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		sum[i]=sum[i-1]+a[i];
	}
	double ans=0.0;
	for(int i=1;i<=n;i++){
		for(int j=i;j<=n;j++){
			ans=ans+(sum[j]-sum[i-1]);
		}
	}
	printf("%.2f\n",ans);
	return 0;
}