hdu 3307 Description has only two Sentences 欧拉定理+快速幂
2013-09-13 21:39
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#include <cstdio> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> using namespace std; #define LL __int64 const LL maxn=1001; LL e[maxn],t; LL gcd(LL a,LL b)//求最大公约数 { return b==0?a:gcd(b,a%b); } LL euler_phi(LL n)//求单个欧拉函数 { LL m=(LL)sqrt(n+0.5); LL i,ans=n; for(i=2;i<=m;i++) if(n%i==0) { ans=ans/i*(i-1); while(n%i==0)n/=i; } if(n>1)ans=ans/n*(n-1); return ans; } void find(LL n)//找出所有因子 { LL m=(LL)sqrt(n+0.5); for(LL i=1;i<m;i++) if(n%i==0){e[t++]=i;e[t++]=n/i;} if(m*m==n)e[t++]=m; } LL pow_mod(LL a,LL b,LL mod)//快速幂 { LL s=1; while(b) { if(b&1) s=(s*a)%mod; a=(a*a)%mod; b=b>>1; } return s; } int main() { LL a,x,y; while(cin>>x>>y>>a) { LL m,phi,i; if(y==0){cout<<"1"<<endl;continue;} m=a/gcd(y/(x-1),a); if(gcd(m,x)!=1){cout<<"Impossible!"<<endl;continue;}//不互质,则x^k%m必定是gcd(m,x)的倍数 phi=euler_phi(m); t=0; find(phi); sort(e,e+t); for(i=0;i<t;i++) { if(pow_mod(x,e[i],m)==1) { cout<<e[i]<<endl; break; } } } return 0; } /* euler_phi(i),欧拉函数,表示求不大于i且与i互质的正整数个数。 本题递推公式化简下可得到通项公式:ak=a0+Y/(X-1)*(X^k-1);后半部分是等比数列的和。 现在求ak%a0=0,即Y/(X-1)*(X^k-1)%a0==0,令m=a0/gcd(Y/(X-1),a0),则可推到求最小的k使得 (X^k-1)%m==0,即X^k==1(mod m). 根据欧拉定理得X^euler_phi(m)==1(mod m).(X与m互质) 又由抽屉原理可知,X^k的余数必定是根据euler_phi(m)的某个因子为循环节循环的。 所以求出最小的因子k使得X^k%m==1,即为答案 */
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