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…

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…

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…

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…

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…

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…

Problem Description Inmany applications very large integers numbers are required. Some of theseapplications are using keys for secure transmission of data, encryption, etc.In this problem you are given a number, you have to determine the number ofdig…

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34084    Accepted Submission(s): 16111 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 ,…