标签：整数 ++ oid size time rom title hat clu
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18742 Accepted Submission(s): 5860
题意：求一个最小的正整数x让2^x mod n = 1.
思路：如果gcd(2,n)!=-1或者n==1的时候显然无解
否则可以用欧拉函数解
欧拉定理：
设gcd(a,m)=1，必有正整数x，使得a^x=1(mod m),且设满足等式的最小正整数为x0，必满足x0|phi（m）.注意m>1.
否则如果gcd(a,m)!=1，则方程a^x=1(mod m)没有解。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; #define LL long long LL e[1000008],t; LL pow_mod(LL a,LL n,LL mod) { LL ans=1; while(n) { if(n&1) ans=ans*a%mod; a=a*a%mod; n>>=1; } return ans; } LL gcd(LL a,LL b) { return b?gcd(b,a%b):a; } LL euler_phi(LL n)//欧拉函数 { LL m=sqrt(n+0.5); LL ans=n,i; for(i=2; i<=m; i++) { if(n%i==0) { ans=ans/i*(i-1); while(n%i==0)n=n/i; } } if(n>1)ans=ans/n*(n-1); return ans; } void finds(LL n) { LL i; e[t++]=n; for(i=2; i*i<=n; i++) { if(n%i==0) { if(i*i==n) e[t++]=i; else { e[t++]=i; e[t++]=n/i; } } } } int main() { int T; LL a,n; while(~scanf("%lld",&n)) { if(n%2==0||n==1) { printf("2^? mod %lld = 1\n",n); continue; } LL m=euler_phi(n); t=0; finds(m); sort(e,e+t); LL ans; for(int i=0; i<t; i++) { if(pow_mod(2,e[i],n)==1) { ans=e[i]; break; } } printf("2^%lld mod %lld = 1\n",ans,n); } return 0; }
标签：整数 ++ oid size time rom title hat clu
原文地址：https://www.cnblogs.com/yi-ye-zhi-qiu/p/9087667.html
