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  • PAT 甲级 1104 sum of Number Segments

    1104. Sum of Number Segments (20)

    时间限制
    200 ms
    内存限制
    65536 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    CAO, Peng

    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) (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 105. 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
    
    
    直接计算每个数出现了几次
    公式:i*(n-i+1)
    #include <iostream>
    #include <string.h>
    #include <algorithm>
    #include <math.h>
    #include <stdio.h>
    #include <stdlib.h>
    
    using namespace std;
    int n;
    double x;
    int main()
    {
        double ans;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%lf",&x);
            ans+=x*i*(n-i+1);
        }
        printf("%.2f
    ",ans);
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/dacc123/p/8228593.html
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