2015 Multi-University Training Contest 1 y sequence
Y sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 667 Accepted Submission(s): 147
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).
Then T cases follow, each contains two positive integer n and r described above.
n<=2*10^18,2<=r<=62,T<=30000.
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int p[] = {, , , , , , , , , , , , , , , , , , };
vector<int>d;
LL n,r;
void init() {
d.clear();
for(int i = ; p[i] <= r; ++i) {
for(int j = d.size()-; j >= ; --j)
if(abs(d[j]*p[i]) <= ) d.push_back(-d[j]*p[i]);
d.push_back(p[i]);
}
}
LL calc(LL x){
if(x == ) return ;
LL ret = x;
for(int i = d.size()-; i >= ; --i){
LL tmp = pow(x+0.5,1.0/abs(d[i])) - ;
if(d[i] < ) ret += tmp;
else ret -= tmp;
}
return ret-;
}
LL solve(){
init();
LL ret = n;
while(true){
LL tmp = calc(ret);
if(tmp == n) break;
ret += n - tmp;
}
return ret;
}
int main() {
ios::sync_with_stdio(false);
int kase;
cin>>kase;
while(kase--){
cin>>n>>r;
cout<<solve()<<endl;
}
return ;
}
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