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  • PAT甲级——1104 Sum of Number Segments (数学规律、自动转型)

    本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90486252

    1104 Sum of Number Segments (20 分)
     

    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

    题目大意:给出N个数,按照要求找出所有的集合,将这些集合的数相加,输出结果。

    思路:找出数学规律,总结公式,需要注意的是,系数k可能会超出int的范围,需要用long long int。(考察了自动转型的知识,题目是Google的人出的)

     1 #include <iostream>
     2 #include <vector>
     3 using namespace std;
     4 
     5 int main()
     6 {
     7     int N;
     8     long long int k;
     9     scanf("%d", &N);
    10     vector <double> v(N);
    11     for (int i = 0; i < N; i++)
    12         scanf("%lf", &v[i]);
    13     double sum = 0;
    14     for (int i = 0; i < N; i++){
    15         k = (i + 1);
    16         k *= N-i;
    17         sum += k * v[i];
    18     }
    19     printf("%.2lf
    ", sum);
    20     return 0;
    21 }
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  • 原文地址:https://www.cnblogs.com/yinhao-ing/p/10914400.html
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