HDU3208 (interval exponent and)

Question: give two numbers a and B, and then each number y in the closed range [a, B] can be expressed as x ^ k = y, requiring x to be as minimal as possible, k is the largest possible, and then calculate the sum of all k. Analysis: For this question, we first need to know that it is calculated based on the following facts: For a number n, from 1 ~ Assume that the number of x in n is in the form of p ^ k, where k can be up to 62. For each k, we need to find that x satisfies the number of x ^ k and is closest to n, now the problem is to find the corresponding x for each k. How can this problem be found? We can consider this. Because we are looking for the nearest number, we can first roughly determine a number r, then we can use r = pow (n, 1/k) for calculation, then we calculate (r-1) ^ k, r ^ k, (r + 1) ^ k, and then we can see which of the three numbers is closest to n, note that (r + 1) ^ k computing may exceed LL, so there are some processing operations. Then, the content is equivalent to the content of the rejection.

#include <iostream>  #include <string.h>  #include <stdio.h>  #include <math.h>    using namespace std;  typedef long long LL;    const LL INF=1e18+300;  const LL T=(LL)1<<31;    LL num[105];    LL multi(LL a,LL b)  {      LL ans=1;      while(b)      {          if(b&1)          {              double judge=1.0*INF/ans;              if(a>judge) return -1;              ans*=a;          }          b>>=1;          if(a>T&&b>0) return -1;          a=a*a;      }      return ans;  }    LL find(LL x,LL k)  {      LL r=(LL)pow(x,1.0/k);      LL t,p;      p=multi(r,k);      if(p==x) return r;      if(p>x||p==-1) r--;      else      {          t=multi(r+1,k);          if(t!=-1&&t<=x) r++;      }      return r;  }    LL Solve(LL n)  {      int i,k=0;      memset(num,0,sizeof(num));      if(n<=3) return n;      num[1]=n;      for(i=2;i<63;i++)      {          num[i]=find(n,i)-1;          if(!num[i]) break;      }      k=i;      for(int i=k-1;i>0;i--)          for(int j=1;j<i;j++)              if(i%j==0) num[j]-=num[i];      LL ans=num[1];      for(int i=2;i<k;i++)          ans+=(i*num[i]);      return ans;  }    int main()  {      LL n,m;      while(cin>>m>>n)      {          if(m==0&&n==0) break;          cout<<Solve(n)-Solve(m-1)<<endl;      }      return 0;  }