2015 Multi-University Training Contest 1 Hdu 5297 Y sequence

最新推荐文章于 2018-09-05 17:15:58 发布

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#多校 #二分 #筛法思想 #反函数

本文探讨了Y序列的性质及求解方法，通过输入正整数n和r，输出Y序列中的第n个数。重点在于筛选排除特定形式的数字，保留剩余的正整数序列。

Y sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 114 Accepted Submission(s): 7

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

二分答案ans，筛[2,ans]前面的a^b的数，判断去掉之后是否等于n。

TLE了。。。pow！！！

据说大表，待更新，，，

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define LL long long
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};
LL T,n,m;
LL input()
{
    char ch;
    LL tmp = 0;
    while(ch=getchar(),ch!='\n'&&ch!=' '){
        tmp=tmp*10+ch-'0';
    }
    return tmp;

}
int main()
{
//    scanf("%lld",&T);
    T = input();
    LL h=1;
    for(int i = 1;i<=62;i++)h=h*2;
    while(T--)
    {
//        scanf("%lld%lld",&n,&m);
        n=input();
        m=input();
//        cout<<n<<" "<<m<<endl;
        LL l = 0,r=n*2;

        //cout<<l<<" "<<r<<endl;
        while(l<r)
        {
            LL mid = (l+r)/2;
            LL tot = 0;
            for(int i = 0;i<25;i++)
                if(prime[i]<=m){
                    LL tmp = //pow(1.0*mid,1.0/prime[i])
                              exp(log(mid)/prime[i]);
                    tot  = tot + tmp - 1;
                    //cout<<i<<"---"<<tmp<<endl;
                }else break;
            //cout<<l<<" "<<r<<" "<<mid<<" "<<tot<<endl;
            if(mid - tot -1>n) r = mid-1;
            else
                if(mid-tot -1==n) r=mid;
            else
                l = mid+1;
        }
        printf("%lld\n",l);
    }
}