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:
|
1 2 |
4 0.1 0.2 0.3 0.4 |
Sample Output:
|
1 |
5.00 |
代码如下：
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
#include<stdio.h> #include<stdlib.h> int main() { double *data,sum=0; int n,i,j; scanf("%d",&n); data=(double *)malloc(n*sizeof(double)); for(i=0;i<n;i++) { scanf("%lf",data+i); } for(i=0;i<n;i++) { sum+=(double)(n-i)*(double)(i+1)*data[i]; } printf("%.2lf\n",sum); } |