HDU 5297 Y sequence

Y sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 658    Accepted Submission(s): 145


Problem Description

Yellowstar likes integers so much that he listed all positive integers in ascending order,but he hates those numbers which can be written as a^b (a, b are positive integers,2<=b<=r),so he removed them all.Yellowstar calls the sequence that formed by the rest
integers“Y sequence”.When r=3,The first few items of it are：

2,3,5,6,7,10......

Given positive integers n and r，you should output Y(n)(the n-th number of Y sequence.It is obvious that Y(1)=2 whatever r is).

 

Input

The first line of the input contains a single number T:the number of test cases.

Then T cases follow, each contains two positive integer n and r described above.

n<=2*10^18,2<=r<=62,T<=30000.

 

Output

For each case,output Y(n).

 

Sample Input

2
10 2
10 3

 

Sample Output

13
14

 

Author

FZUACM

 

Source

2015 Multi-University Training Contest 1

 

#include <bits/stdc++.h>
using namespace std;
#define prt(k) cerr<<#k" = "<<k<<endl
typedef long long ll;
const int N = 63;
int p
;
ll n; int r;
int sign
;
vector<int> rc;
vector<int> mi;
void get_rc()
{
rc.clear();
for (int x : mi) {
int n = rc.size();
if (abs(x) > r) break;
for (int j=0;j<n;j++)
if (abs(x*rc[j]) <= 62)
rc.push_back(x*rc[j]);
rc.push_back(x);
}
}
ll f(ll n, int r)
{
if (n==1) return 0;
ll ans  = n - 1;
for (ll x:rc) {
ll t = pow(n+0.5, 1.0/abs(x) ) - 1;
if (x < 0)
ans -= t;
else
ans += t;
}
return ans ;
}
int main()
{
for (int i=1;i<N;i++) p[i] = i;
for (int i=2;i<N;i++) if (p[i]==i) {
for (int j=i+i; j<N; j+=i) p[j] = i;
mi.push_back(-i);
}
int re, ca=1;
scanf("%d", &re);
while (re--) {
scanf("%I64d%d", &n, &r);
get_rc();
ll x = n ;
while (1) {
ll t = f(x, r);
if (t >= n) break;
x += n - t;
}
printf("%I64d\n", x);
}
return 0;
}