hdu-1395 2^x mod n = 1
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2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11542 Accepted Submission(s): 3577
Problem Description
Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.
Input
One positive integer on each line, the value of n.
Output
If the minimum x exists, print a line with 2^x mod n = 1.
Print 2^? mod n = 1 otherwise.
You should replace x and n with specific numbers.
Print 2^? mod n = 1 otherwise.
You should replace x and n with specific numbers.
Sample Input
25
Sample Output
2^? mod 2 = 12^4 mod 5 = 1
Author
MA, Xiao
Source
ZOJ Monthly, February 2003
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//跟上一题差不多。这题直接告诉底数为2 求最小的满足2^xmod n==1的最小x,解法是一样 的。
#include<iostream>#include<cstdio>#include<cmath>#include<cstring>using namespace std;const int Max = 1000010;int prime[Max], phi[Max]; //保存所有值的欧拉函数void fun() //求1到max所有值的欧拉函数{ int i,j; prime[0] = prime[1] = 0; for(i = 2;i <= Max; i ++) prime[i]=1; for(i = 2; i*i <= Max; i ++) if(prime[i]) for(int j=i*i;j<=Max;j+=i) prime[j]=0; for(i=1;i<=Max;i++) phi[i]=i; for(i=2;i<=Max;i++) if(prime[i]) for(j = i; j <= Max; j += i) phi[j] = phi[j]/i * (i-1);}int Mod(int a, int b, int c) //快速幂取模 { int ans = 1; __int64 aa = a; while(b) { if (b % 2) ans = ans * aa % c; aa = aa * aa % c; b /= 2; } return ans;}int main(){// freopen("a.txt","r",stdin);// freopen("b.txt","w",stdout); fun(); int a, n,i; while(scanf("%d",&n)!=EOF) { int sn = (int)sqrt(phi[n]), ans = n; //在1-sn之间枚举n的欧拉函数的因子 for(i = 1; i <= sn; i++) //在欧拉函数的所有因子中查找满足条件并且是最小的。 { if (phi[n] % i == 0) //如果i是phi的因子 更新最小值 { if (Mod(2, i, n) == 1 && ans > i) ans = i; if (Mod(2, phi[n] / i, n) == 1 && ans > phi[n] / i) ans = phi[n] / i; } } if(ans==n) printf("2^? mod %d = 1\n",n); else printf("2^%d mod %d = 1\n",ans,n); //cout << ans << endl; } return 0;}
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