题目链接:POJ 3641 Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, know…

http://poj.org/problem?id=3641 练手用,结果念题不清,以为是奇偶数WA了一发 #include<iostream> #include<cstdio> #include<cmath> using namespace std; typedef long long ll; bool judge_prime(ll k) { ll i; ll u=int(sqrt(k*1.0)); ;i<=u;i++) { ) ; } ; } ll mod_p…

题目连接 http://poj.org/problem?id=3641 Pseudoprime numbers Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but…

题目链接 题意:题目定义了Carmichael Numbers 即 a^p % p = a.并且p不是素数.之后输入p,a问p是否为Carmichael Numbers? 坑点:先是各种RE,因为poj不能用srand()...之后各种WA..因为里面(a,p) ?= 1不一定互素,即这时Fermat定理的性质并不能直接用欧拉定理来判定..即 a^(p-1)%p = 1判断是错误的..作的 #include<iostream> #include<cstdio> #include&l…

传送门:http://poj.org/problem?id=3461 题目大意:给你两个字符串p和s,求出p在s中出现的次数. 题解:这一眼看过去就知道是KMP,作为模板来写是最好不过了.... 这道题我写了两种风格的kmp,个人感觉第2种好理解一些; #include <iostream> #include <cstring> #include <cstdio> #define inf ],p[]; ]; int lens,lenp,n; using namespac…

http://poj.org/problem?id=3461 直接KMP就好.水题 #include<cstdio> #include<cstring> const int MAXN=10000+10; const int MAXM=1000000+10; char P[MAXN],T[MAXM]; int f[MAXN],n,m,ans; void getFail() { f[0]=f[1]=0; for(int i=1;i<n;i++){ int j = f[i]; wh…

 强伪素数 题目大意:利用费马定理找出强伪素数(就是本身是合数,但是满足费马定理的那些Carmichael Numbers) 很简单的一题,连费马小定理都不用要,不过就是要用暴力判断素数的方法先确定是不是素数,然后还有一个很重要的问题,那就是a和p是不互质的,不要用a^(p-1)=1(mod p)这个判据,比如4^6=4(mod 6),但是4^5=4(mod 6) #include <iostream> #include <functional> #include <algo…

Pseudoprime numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6044   Accepted: 2421 Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power…

#include <cstring> #include <cstdio> #include <iostream> #include <cmath> #include <algorithm> using namespace std; #define LL long long LL p,res,a; bool judge_prime(LL k) { LL i; LL u=int(sqrt(k*1.0)); ;i<=u;i++) { ) ; }…

Description Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a ps…