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  • PAT甲级——A1104 Sum of Number Segments

    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 }
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  • 原文地址:https://www.cnblogs.com/zzw1024/p/11441416.html
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