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

    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 }
     
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  • 原文地址:https://www.cnblogs.com/bianjunting/p/13211081.html
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