#include#include#define ll long longll mod;bool Judge(int x){ for(int i=2;i0) { if(n&1) { ans*=ret; ans%=mod; } ret=(ret*ret)%mod; n>>=1; } return ans;}int main(){ int p,a; while(scanf("%d %d",&p,&a)!=EOF) { if(p==0...
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2013-09-26 16:21:00
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HPU专题训练（-1）GCD&&素筛&&快速幂_____K - Pseudoprime numbers题目来源http://poj.org/problem?id=3641第一次做的时候用的是int 没有用long long 范围比较小，出错了而且英文题目，还扯了费马小定理，百度了一下，但是也没用到写两个函数就行了，快速幂，素数判断/*...
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2022-10-18 16:46:41
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Pseudoprime numbersTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 4864Accepted: 1872DescriptionFermat'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 ve
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2021-07-29 16:17:46
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ed: 5367 DescriptionFermat's theorem states that for a...
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2023-02-07 12:13:36
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题 Description Jerry is caught by Tom. He was penned up in one room with a door, which only can be opened by its code. The code is the answer of the su
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2021-07-22 14:02:31
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#include <iostream> //二分using namespace std;__int64 p;bool prime(__int64 n){ for(__int64 i=2;i*i<=n;++i) if(n%i==0) return false; return true;}__int64 query(__int64 a,__int64 n){ if(n==1) return a; __int64 q=query(a,n/2),t=(q*q)%p; if(n%2==0) return t; else return (t*a)%p;}int main(){ __int
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2011-07-22 20:19:00
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#include<iostream> #include<stdio.h>#include<cmath>using namespace std;int is_prime(int x){ int m=floor(sqrt((double)x)+0.5); for(int i=2;i<=m;++i) if(x%i==0) return 0; return 1;}int power(int a,int b,int m) //快速幂取模计算 (a^b)%m{ if(a==0) return 0; else if(...
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2011-07-05 02:00:00
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【解题思路】这是一道不错的数论题目，可以学到很多东西，比如说将指数很大的数进行二进制处理，还有就是判断素数和伪素数。最有用的知识点是：1、如果p是奇数，则有(a.^p)mod(m)==((a%m)*(a.^(p-1)%m)%m成立；2、如果p是偶数，则有a.^p==a.^(p/2)*a.^(p/2)。也许可以说这两个结论人人都知道，但用起来就不是那么一回事了。#
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2022-08-05 15:48:11
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Pseudoprime numbers
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 11336
Accepted: 4891
Description
Fermat's theorem states that for any prime number p and for any integer a >
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2021-08-31 10:43:34
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题意：a的p次方p取模等于a，且p不是素数，就输出yes；#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#define ll long longusing namespace std;ll powermod(ll a
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2022-10-19 16:13:14
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Pseudoprime numbers
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 11336
Accepted: 4891
Description
Fermat's theorem states that for any prime number p and for any integer a &g
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2021-08-31 10:43:33
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DescriptionFermat'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 ...
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2015-11-09 20:41:00
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Fermat's theorem states
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2023-05-18 14:19:07
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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 ...
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2023-05-09 10:03:19
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题目链接：http://poj.org/problem?id=3641 题目大意：给定a,p，判断一个数是不是伪素数（伪素数是指p---p满足a^p = a (mod p)但p本身不是素数），2>= 1; } return ret; } POJ 3641代码：#i...
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2013-01-12 21:24:00
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快速幂：快速幂就是所求的幂次方过大，导致代码所用的时间超限。如：求2^3,3的二进制是11，(n&1)判断次方数的二进制是否为1，n>>1,向右进位1：代码：k=1,t=n;while(n) { if(n&1)//判断n的最后一位二进制不为0 { k=k*m; } n=n>>1; m=m*m; }题目描述：Ferm...
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2023-02-14 16:21:18
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Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld
& %llu
Submit Status
Description
Fermat's theorem states that for any prime number p and for any integer a > 1, a
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2022-08-24 11:39:10
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http://poj.org/problem?id=3641注意计算前先 a%=p 一下。/*16ms,384KB*/#includelong long qmod(long long a, long long n, long long b){ long long ans = 1; while (n) { if (n & 1) { ans = (ans
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2023-04-12 03:25:50
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DescriptionFermat'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
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2022-05-27 17:11:05
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题目大意：费马
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2014-09-23 06:51:17
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