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 1. 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
这就是一道找规律题
每个数字出现的次数为(i+1)*(n-i)次
1 #include <iostream> 2 #include <vector> 3 using namespace std; 4 int n; 5 double sum = 0.0; 6 int main() 7 { 8 cin >> n; 9 double *v = new double[n]; 10 for (int i = 0; i < n; ++i) 11 cin >> v[i]; 12 for (int i = 0; i < n; ++i) 13 sum += v[i] * (i + 1)*(n - i); 14 printf("%.2f ", sum); 15 return 0; 16 }