PAT_A1104#Sum of Number Segments

Source:

PAT A1104 Sum of Number Segments (20 分)

Description:

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

Key:

• 模拟题

Attention:

• i*(n-i+1)*x，3和4测试点会报错

Code:

 1 #include<cstdio>
 2 
 3 int main()
 4 {
 5 #ifdef ONLINE_JUDGE
 6 #else
 7     freopen("Test.txt", "r", stdin);
 8 #endif // ONLINE_JUDGE
 9 
10     int n;
11     double x,sum=0;
12     scanf("%d", &n);
13     for(int i=1; i<=n; i++)
14     {
15         scanf("%lf", &x);
16         sum += (x*i*(n-i+1));
17     }
18     printf("%.2f\n", sum);
19 
20     return 0;
21 }

PAT_A1104#Sum of Number Segments

原文：https://www.cnblogs.com/blue-lin/p/11385629.html