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…

LINK:HDU 1018 题意:求n!的位数~ 由于n!最后得到的数是十进制,故对于一个十进制数,求其位数可以对该数取其10的对数,最后再加1~ 易知:n!=n*(n-1)*(n-2)*......*3*2*1 ∴lg(n!)=lg(n)+lg(n-1)+lg(n-2)+......+lg(3)+lg(2)+lg(1); 代码: #include <iostream> #include <cstdio> #include <cmath> using namespace…

[问题描述] 数根问题递归求解:输入n个正整数(输入格式中第一行为整数个数n,后续行为n个整数),输出各个数的数根.数根的定义:对于一个正整数n,我们将它的各个位相加得到一个新的数字,如果这个数字是一位数,我们称之为n的数根,否则重复处理直到它成为一个一位数,这个一位数也算是n的数根.例如:考虑24,2+4=6,6就是24的数根.考虑39,3+9=12,1+2=3,3就是39的数根.? 要求计算一个数的数根部分利用递归函数实现. 样例输入: 5 23 424 98 632 12345 样例输出:…

Big Number 题意:算n!的位数. 题解:对于一个数来算位数我们一般都是用while去进行计算,但是n!这个数太大了,我们做不到先算出来在去用while算位数. while(a){ cnt++; a/=; } 将一个数对取10对数(取整),然后再加一就是这个数的位数,然后我们在算n!的时候每次对10取对数就好了. #include<iostream> #include<cmath> using namespace std; int main() { ios::sync_wi…

Description 根据密码学需要,要计算某些数的阶乘的位数. Input 第一行为整数n ,接下来 n 行, 每行1个数m (1 ≤ m ≤ 10^7) . Output 输出m的阶乘的位数. Sample Input 2 10 20 Sample Output 7 19 思路:求n的位数:(int)log10(n)+1, log(a*b)=log(a)+log(b) #include <cstdio> #include <cmath> using namespace std…

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21106    Accepted Submission(s): 9498 Problem Description In many applications very large integers numbers are required. Some of these…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1018 解题报告:输入一个n,求n!有多少位. 首先任意一个数 x 的位数 = (int)log10(x) + 1; 所以n!的位数 = (int)log10(1*2*3*.......n) + 1; = (int)(log10(1) + log10(2) + log10(3) + ........ log10(n)) + 1; #include<cstdio> #include<cstrin…

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 40262    Accepted Submission(s): 19637 Problem Description In many applications very large integers numbers are required. Some of these…

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…

题意: 给一个数n,返回该数的阶乘结果是一个多少位(十进制位)的整数. 思路: 用对数log来实现. 举个例子 一个三位数n 满足102 <= n < 103: 那么它的位数w 满足 w = lg103 = 3. 因此只要求lgn 向下取整 +1就是位数.然后因为阶乘比如5阶乘的话是5 * 4 * 3 * 2 * 1.位数就满足lg 5 * 4 * 3 * 2 * 1 = lg5 + lg4 + lg3 + lg2 + lg1.用加法就不会超过数字上限. 当然这是十进制下得.如果是m进制下 ,…