ACdream1084

来源：互联网发布：怀化天乙网络编辑：程序博客网时间：2024/06/05 16:47

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题意：将A(n , m)转化成k进制后末尾0的个数

代码：

#include <iostream>#include <cstring>using namespace std;long long prime[500005],num[500005];long long fac[50005],count[50005];long long getsum(long long a,long long b){    long long sum=0;    while(a){    a/=b;    sum+=a;    }    return sum;}int main(){   long long i,j,k,t,n,m,x,temp,minn,sumsum;   k=0;   memset(prime,0,sizeof(prime)) ;   for(i=2;i<=500000;i++){   if(!prime[i])   num[k++]=i;   for(j=0;j<k&&num[j]*i<=500000;j++){   prime[num[j]*i]=1;   if(i%num[j]==0)   break;   }   }   cin>>t;   while(t--){   memset(count,0,sizeof(count));   cin>>n>>m>>x;                      //转换成k进制有多少个零相当于能连续除几个k   temp=0;                            //因此将k进行素因子分解，例如20=2*2*5。。。    for(i=0;i<k&&num[i]*num[i]<=x;i++){//阶乘有几套k的素因子就有几个零   if(x%num[i]==0){   fac[temp]=num[i];   while(x%num[i]==0){   x/=num[i];   count[temp]++;   }   temp++;   }   }   if(x>1){   fac[temp]=x;count[temp]=1;   temp++;   }//   for(i=0;i<temp;i++)//    cout<<fac[i]<<" "<<count[i]<<endl;   minn=9999999999;   for(i=0;i<temp;i++){   sumsum=getsum(n,fac[i])-getsum(n-m,fac[i]);   minn=min(sumsum/count[i],minn);   }   cout<<minn<<endl;   }   return 0;}</span>

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