如果有相应的OJ题目,欢迎同学们提供相应的链接 相关链接 所有模板的快速链接 扩展欧几里得extend_gcd模板 poj_1006_Biorhythms,my_ac_code 简单的测试 None 代码模板 /* * TIME COMPLEXITY:O(nlogm) * PARAMS: * a x==ai(mod mi) * m * n the number of equation. */ int crt(int a[],int m[],int n){ int M=1; for(int i=0…

int exgcd(int a,int b,int &x,int &y) { ) { x=,y=; return a; } int gcd=exgcd(b,a%b,x,y); int t=x; x=y; y=t-a/b*y; return gcd; } int China(int W[],int B[],int k) { ,m,n=; ;i<=k;i++) n*=W[i]; ;i<=k;i++) { m=n/W[i]; exgcd(W[i],m,x,y); a=(a+y*m*B…

中国剩余定理 CRT 正常版本CRT 要解的是一个很容易的东西 \[ \begin{aligned} x\equiv a_1(mod\ m_1)\\ x\equiv a_2(mod\ m_2)\\ ...\\ x\equiv a_n(mod\ m_n) \end{aligned} \] 保证\(m_1,m_2...m_n\)之间两两互质,求最小的\(x\). 设\(M=\prod m_i\). 首先我们确定一点,我们求出了任意一个满足条件的\(x\)之后,只需要对其模\(M\)就是最终的答案.…

Hello Kiki Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 247 Accepted Submission(s): 107   Problem Description One day I was shopping in the supermarket. There was a cashier counting coins serio…

http://poj.org/problem?id=2891 Strange Way to Express Integers Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 11970   Accepted: 3788 Description Elina is reading a book written by Rujia Liu, which introduces a strange way to express no…

http://acm.hdu.edu.cn/showproblem.php?pid=1573 X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4439    Accepted Submission(s): 1435 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0],…

X问题 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3921    Accepted Submission(s): 1253 Problem Description 求在小于等于N的正整数中有多少个X满足:X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod…

Biorhythms Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 103539   Accepted: 32012 Description Some people believe that there are three cycles in a person's life that start the day he or she is born. These three cycles are the physical,…

中国剩余定理的非互质形式 任意n个表达式一对对处理,故只需处理两个表达式. x = a(mod m) x = b(mod n) km+a = b (mod n) km = (a-b)(mod n) 利用扩展欧几里得算法求出k k = k0(mod n/(n,m)) = k0 + h*n/(n,m) x = km+a = k0*m+a+h*n*m/(n,m) = k0*m+a (mod n*m/(n,m)) #include <cstdio> #include <cstring> #…

Two Arithmetic Progressions 题目链接: http://codeforces.com/contest/710/problem/D Description You are given two arithmetic progressions: a1k + b1 and a2l + b2. Find the number of integers x such that L ≤ x ≤ R and x = a1k' + b1 = a2l' + b2, for some inte…