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  • HDU 1018 Big Number

    Big Number

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 38814    Accepted Submission(s): 18850


    Problem Description
    In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of digits in the factorial of the number.
     
    Input
    Input consists of several lines of integer numbers. The first line contains an integer n, which is the number of cases to be tested, followed by n lines, one integer 1 ≤ n ≤ 107 on each line.
     
    Output
    The output contains the number of digits in the factorial of the integers appearing in the input.
     
    Sample Input
    2 10 20
     
    Sample Output
    7 19
     
    Source
     
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    分析:  对于一个数n  它的位数等于 log10(n)+1;
                那么n! 的位数为 log10(n*(n-1).....*1) +1
              公式展开  log10(n)+log10(n-1)+....log10(1)+1;
    代码如下:
    #include <cstdio>
    #include <iostream>
    #include <cstring>
    #include <map>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    typedef long long ll;
    const int MAXN=40000;
    int ans[MAXN];
    int flag;
    int main()
    {
        double sum;
        int t,n;
        scanf("%d",&t);
        while(t--)
        {
            sum=1;
            scanf("%d",&n);
            for(int i=1;i<=n;i++)
            {
              sum+=log10(i);
            }
            printf("%d
    ",(int)sum);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/a249189046/p/7487769.html
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