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

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

 

解题：传说中的容斥原理。

 

先计算1到n间有多少个数被删除了，那么我们就还需要m=n+删除的数目，看看又又多少个删除了，看看剩下的是不是刚好n个，否则补上n - 剩下的个数，继续搞

 

处理1的时候，先把1都不算，最后才算，奇数个素因子的乘积那么要加上，偶数个素因子的乘积要减去

 

参考了这位博主的写法

 

可以由$\sqrt[i]{a}$得出范围内指数是i的元素的个数  

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int p[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67};
vector<int>d;
LL n,r;
void init() {
    d.clear();
    for(int i = 0; p[i] <= r; ++i) {
        for(int j = d.size()-1; j >= 0; --j)
            if(abs(d[j]*p[i]) <= 63) d.push_back(-d[j]*p[i]);
        d.push_back(p[i]);
    }
}
LL calc(LL x){
    if(x == 1) return 0;
    LL ret = x;
    for(int i = d.size()-1; i >= 0; --i){
        LL tmp = pow(x+0.5,1.0/abs(d[i])) - 1;
        if(d[i] < 0) ret += tmp;
        else ret -= tmp;
    }
    return ret-1;
}
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 0;
}