时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CAO, Peng
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) (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 105. 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
直接计算每个数出现了几次
公式:i*(n-i+1)
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
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;
}