1104. Sum of Number Segments (20)

1104. Sum of Number Segments (20)

 

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

 

 

#include <stdio.h>

int main()
{
double sum=0,num;
int N,i;
scanf("%d",&N);
for(i=0;i<N;i++)
{
scanf("%lf",&num);
sum+=num*(N-i)*(i+1);
}
printf("%.2lf\n",sum);
return 0;
}