1104 Sum of Number Segments (20分)(long double)

Posted 2020-12-13 bianjunting

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

long double的输出方式为 "%Lf"

 1 #include <cstdio>
 2 using namespace std;
 3 
 4 const int maxn = 1e5 + 5;
 5 int n;
 6 double val[maxn];
 7 
 8 int main() {
 9     scanf("%d", &n);
10     for(int i = 0; i < n; i ++) {
11         scanf("%lf", &val[i]);
12     }
13     long double sum = 0;
14     for(int i = 0; i < n; i ++) {
15         sum += val[i] * (i + 1) * (n - i);
16     }
17     printf("%.2Lf
", sum);
18     return 0;
19 }