Poj3696 The Lukiest Number

传送门

Solution

懒得写啦

Code

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #define Rg register
 #define go(i,a,b) for(Rg int i=a;i<=b;i++)
 #define yes(i,a,b) for(Rg int i=a;i>=b;i--)
 #define ll long long
 using namespace std;
 int read()
 {
     int x=,y=;char c=getchar();
     while(c<''||c>''){if(c=='-')y=-;c=getchar();}
     while(c>=''&&c<=''){x=(x<<)+(x<<)+c-'';c=getchar();}
     return x*y;
 }
 ll T,L,d,n,m,ans,c;
 bool flag=;
 ll gcd(ll x,ll y){return y==?x:gcd(y,x%y);}
 ll phi(ll x)
 {
     ll sm=x,y=x;
     go(i,,sqrt(y))
     {
         if(x%i==)sm=sm/i*(i-);
         while(x%i==)x/=i;
     }
     if(x>)sm=sm/x*(x-);
     return sm;
 }
 ll mul(ll x,ll y,ll mod)
 {
     ll sm=;
     while(y)
     {
         if(y&)sm=(sm+x)%mod;
         x=(x<<)%mod;y>>=;
     }
     return sm;
 }
 ll ksm(ll x,ll y,ll mod)
 {
     ll sm=;
     while(y)
     {
         if(y&)sm=mul(sm,x,mod);//注意这里直接乘起来会溢出
         x=mul(x,x,mod);y>>=;
     }
     return sm;
 }
 int main()
 {
     while()
     {
         scanf("%lld",&L);if(!L)break;T++;flag=;ans=;
         if(%L==){printf("Case %lld: 1\n",T);continue;}
         d=gcd(L,);n=*L/d;m=phi(n);c=sqrt(m);
         go(i,,c)
         {
             if(m%i==&&ksm(,i,n)==){ans=i;flag=;break;}
         }
         if(!flag)
             yes(i,c,)
             {
                 if(m%i==&&ksm(,m/i,n)==){ans=m/i;break;}
             }
         printf("Case %lld: %lld\n",T,ans);
     }
     return ;
 }