poj3358

• #include<iostream>#include<cstdlib>#include<cstdio>using namespace std;typedef long long LL;LL pow(LL a,LL p,LL m){LL d=1,t=a;while(p>0){   if(p&1==1){     d=(d*t)%m;      }      t=(t*t)%m;      p=p>>1;}return d;}int  enlerFun(int n){int count = n;    for(int i=2;i<=n;i++){      if(n%i==0){       count=count-count/i;       while(n%i==0){n=n/i;}       }       }    return count;}LL gcd(LL a,LL b){ return b==0?a : gcd(b,a%b);}int main(){      int t=0;    LL p,q,d,time,yue,minyue; //minyue最小的能使公式成立的约数     while( ~ scanf("%I64d/%I64d",&p,&q)  ){            d=gcd(p,q);p/=d;q/=d;  //化简分子和分母       //printf("%I64d %I64d",p,q);       time=1;     while( (q&1)==0 ){  //注意添加括弧               time++;              q=q>>1;             }     //time就是起始位置3088601       yue=enlerFun(q);     minyue=yue;     //printf("%d\n",minyue);                    for(int i=1;i*i<=yue;i++){        if(yue%i==0){            if(pow(2,i,q)==1 &&  i<minyue){minyue=i;} //符合公式，大擂台找最小             if(pow(2,yue/i,q)==1 &&  (yue/i)<minyue){minyue=(yue/i);}        }             //printf("***%d\n",minyue);     }     printf("Case #%d: %d,",++t,time); printf("%d\n",minyue);    }//whilereturn 0;}