题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3123 The GNU Compiler Collection (usually shortened to GCC) is a compiler system produced by the GNU Project supporting various programming languages. But it doesn’t contains the math operator “!”.In mat…

GCC Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 3867    Accepted Submission(s): 1272 Problem Description The GNU Compiler Collection (usually shortened to GCC) is a compiler system produc…

这题分2种情况: 1) n>=m时,k!%m=0(k>=m),所以只需令n=m-1即可: 2) n<m时,正常情况处理即可. ;}…

GCC Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 3754    Accepted Submission(s): 1216 Problem Description The GNU Compiler Collection (usually shortened to GCC) is a compiler system produce…

Problem Description A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer. For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number. Now you should find out all the DFS numbe…

题目 输入:n 和 mod 输出: Output the answer of (0! + 1! + 2! + 3! + 4! + ... + n!)%m. Constrains 0 < T <= 20 0 <= n < 10^100 (without leading zero) 0 < m < 1000000 一道模运算的题,其实是一道数学题.因为n太大了,所以用字符串来存. 当n大于m时 ,n的阶乘中必定包含因数m,所以取余后必定为0.所以尽管数很大,但是只用求到这个…

B - 2 Time Limit:5000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1042 Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!  Input One N in one line, process to the end of file.  Out…

以为有啥牛逼定理,没推出来,随便写写就A了----题非常水,可是wa了一次 n>=m  则n!==0 注意的一点,最后 看我的凝视 #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <iostream> using namespace std; const int maxn = 115; #define ll long lo…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1018 Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 42715    Accepted Submission(s): 20844 Problem Description In many applications very…

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 ,…