标签：PAT int Sum 0.2 Number 0.4 0.1 sum 0.3
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\n", sum);
20 return 0;
21 }
标签：PAT,int,Sum,0.2,Number,0.4,0.1,sum,0.3 来源： https://www.cnblogs.com/yinhao-ing/p/10914400.html
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