题目描述

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.

输入描述

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.

输出描述

For each test case, print in one line the sum of all the numbers in all the segments, accurate up to 2 decimal places.

输入例子

4
0.1 0.2 0.3 0.4

输出例子

5.00

我的代码

//正常思路，运行超时 
#include<stdio.h>
int main(){
    double a[100001];
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%lf",&a[i]);
    }
    double sum=0,ans=0;
    for(int i=0;i<n;i++){
        for(int j=i;j<n;j++){
            sum=sum+a[j];
            ans=ans+sum;
        }
        sum=0;
    }
    printf("%.2f",ans);
    return 0;
}

//发现规律后的解答
#include<stdio.h>
#include<stdio.h>
int main(){
    double a,count;
    double n;
    scanf("%lf",&n);
    double ans=0;
    for(int i=1;i<=n;i++){
        scanf("%lf",&a);
        count=i*(n+1-i);
        ans=ans+a*count;
    }
    printf("%.2f",ans);
    return 0;
}