Y sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1192 Accepted Submission(s): 265
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
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Mobius函数+迭代
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXT (30000+10)
#define MAXR (63)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
const long long prime[] = {2,3,5,7,11,13,17,19,23,29,
31,37,41,43,47,53,59,61,67,
71,73,79,83,89,97};
class Math
{
public:
ll gcd(ll a,ll b){if (!b) return a;return gcd(b,a%b);}
ll abs(ll x){if (x>=0) return x;return -x;}
ll exgcd(ll a,ll b,ll &x, ll &y)
{
if (!b) {x=1,y=0;return a;}
ll g=exgcd(b,a%b,x,y);
ll t=x;x=y;y=t-a/b*y;
return g;
}
ll pow2(ll a,int b,ll p) //a^b mod p
{
if (b==0) return 1;
if (b==1) return a;
ll c=pow2(a,b/2,p);
c=c*c%p;
if (b&1) c=c*a%p;
return c;
}
ll Modp(ll a,ll b,ll p) //a*x=b (mod p)
{
ll x,y;
ll g=exgcd(a,p,x,y),d;
if (b%g) {return -1;}
d=b/g;x*=d,y*=d;
x=(x+abs(x)/p*p+p)%p;
return x;
}
ll mobius(ll x)
{
}
}S;
ll n;
int r;
int num[600000],Nnum,mobnum[600000],rNnum[600000];
void get() {
Nnum=0;
for(int i=0;prime[i]<=MAXR;i++) {
int m=Nnum;
For(j,m) {
if (prime[i]*num[j]<=MAXR) {
num[++Nnum]=prime[i]*num[j];
mobnum[Nnum]=-mobnum[j];
}
}
num[++Nnum] = prime[i];
mobnum[Nnum]=-1;
rNnum[i]=Nnum;
}
}
ll calc(ll x) {
ll ans=0;
int rN=0;
while (prime[rN+1]<=r) ++rN;
For(i,rNnum[rN])
{
ans+=-mobnum[i]*(ll)(pow(x+0.1,1.0/num[i])-1);
}
return x-ans-1;
}
ll Y(ll n,int r)
{
ll ans=n;
while(1) {
ll tmp=calc(ans);
if (tmp==n) break;
ans+=n-tmp;
}
return ans;
}
int main()
{
// freopen("J.in","r",stdin);
// freopen(".out","w",stdout);
get();
// For(i,Nnum) cout<<num[i]<<' '<<mobnum[i]<<endl;
int T;
scanf("%d",&T);
while(T--) {
scanf("%lld%d",&n,&r);
printf("%lld\n",Y(n,r));
}
return 0;
}
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