*乘号的笔记本

http://acm.hdu.edu.cn/showproblem.php?pid=1087Problem DescriptionNowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy

参考过程：点击打开链接参考代码：点击打开链接康托展开表示的是当前排列在n个不同元素的全排列中的名次。比如213在这3个数所有排列中排第3。 那么，对于n个数的排列，康托展开为： 其中表示第i个元素在未出现的元素中排列第几。举个简单的例子： 对于排列4213来说，4在4213中排第3，注意从0开始，2在213中排第1，1在13中

1.std::next_permutation函数原型template class BidirectionalIterator>　　bool next_permutation (BidirectionalIterator first, BidirectionalIterator last );　　template class BidirectionalIterator, class

二分算a的n次方：int pow(int a,int n){ int ans=1; while(n) { if(n&1) ans*=a; a*=a; n>>=1; } return ans;}快速幂：b and 1 //也就是取b的二进制最低

http://acm.hdu.edu.cn/showproblem.php?pid=3037Problem DescriptionAlthough winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get th

Chinese remainder theorem(CRT):中国剩余定理是求解一次线性同余方程组的方法。中国剩余定理： 假设整数m1,m2, … ,mn两两互素，则同余方程组  有整数解。 设 是m1,m2,m3…mn的乘积，并设 是除了mi以外的n-1个整数的乘积。 设是Mi模mi的逆元（数论中的倒数）。那么在模M下的解是唯一的。

http://poj.org/problem?id=1061   A - 青蛙的约会 POJ - 1061 #include#includeusing namespace std;long long x,y,q;long long X,Y,M,N,L;void exgcd(long long a,long long b){ if(b==0)

Striling数Bell数复习：http://blog.youkuaiyun.com/acdreamers/article/details/12309269http://blog.youkuaiyun.com/creat2012/article/details/40619031（模板）Bell(hdu4767+矩阵+中国剩余定理+bell数+欧几里德)

Problem DescriptionIn 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

思路：设(A/B)%9973 = K, 则A/B = k + 9973x  (x未知）， 因此A = kB + 9973xB,又A%9973 = n， 所以kB%9973 = n,  故kB = n + 9973y (y未知)故(k/n)B +(-y/n)*9973 = gcd(B,9973) = 1扩展欧几里得 求出k/n,  再乘以个n，记得取模，就是answer了

求最大公约数的线性组合（扩展欧几里得定理）：#include #include #include #include #include #include #include using namespace std;int x,y;int gcd(int a,int b){ //欧几里得算法求最大公约数 if(b==0)

https://vjudge.net/problem/POJ-2407RelativesTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14310 Accepted: 7151DescriptionGiven n, a positi

http://acm.hdu.edu.cn/showproblem.php?pid=2095Problem DescriptionIn the new year party, everybody will get a "special present".Now it's your turn to get your special present, a lot of pr

Problem B: Zhazhahe究竟有多二DescriptionZhazhahe竟然能二到把耳机扔到洗衣机里去洗，真的是二到了一种程度，现在我们需要判断一下zhazhahe二的程度（就是计算zhazhahe的脑残值有几个2的因子），下面给你一个n，n!表示zhazhahe的脑残值。Input输入一个正整数t(0表示样例组数，每组样例输入一个正整

http://acm.hdu.edu.cn/showproblem.php?pid=1061Problem DescriptionGiven a positive integer N, you should output the most right digit of N^N. InputThe input contains severa

http://poj.org/problem?id=2773DescriptionTwo positive integers are said to be relatively prime to each other if the Great Common Divisor (GCD) is 1. For instance, 1, 3, 5, 7, 9...are all

http://poj.org/problem?id=1284DescriptionWe say that integer x, 0 i mod p) | 1 <= i <= p-1 } is equal to { 1, ..., p-1 }. For example, the consecutive powers of 3 modulo 7 are 3, 2, 6, 4

对应练习题：SDNUOJ1 1286点击打开链接1.Miller-rabin算法：Miller-rabin算法是一个用来快速判断一个正整数是否为素数的算法。根据费马小定理，如果p是素数，则a^(p-1)≡1(mod p)对所有的a∈[1,n-1]成立。所以如果在[1,n-1]中随机取出一个a，发现不满足费马小定理，则证明n必为合数。【但是每次