hdu 5297 容斥原理

很容易想到我们需要这样一个函数f(n)表示的是1~n的数字中在y sequence中的个数，于是可以想到用容斥来做，先假设答案是n然后计算n中y sequence的个数，然后n加上不够的，继续判断，一直迭代求出答案。

小技巧：预处理的时候素数设为负数方便判断容斥的时候是应该加还是减。

#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <cmath>
using namespace std;

typedef long long ll;
int p[] = {-2,-3,-5,-7,-11,-13,-17,-19,-23,-29,-31,-37,-41,-43,-47,-53,-59,-61,-67};
vector<int> v;

void init( int r )
{
    v.clear();
    for ( int i = 0; -p[i] <= r; i++ )
    {
        int bound = v.size();
        for ( int j = 0; j < bound; j++ )
        {
            if ( abs( v[j] * p[i] ) <= 63 )
            {
                v.push_back( v[j] * p[i] );
            }
        }
        v.push_back(p[i]);
    }
}

ll cal( ll n )
{
    ll res = 0;
    for ( int i = 0; i < v.size(); i++ )
    {
        ll tmp = pow( n + 0.5, 1.0 / abs(v[i]) ) - 1;
        if ( v[i] < 0 ) res += tmp;
        else res -= tmp;
    }
    return n - res - 1;
}

int main ()
{
    int t;
    scanf("%d", &t);
    while ( t-- )
    {
        ll n, cp, tmp;
        int r;
        scanf("%I64d %d", &n, &r);
        init(r);
        cp = n, tmp = cal(cp);
        while ( tmp < n )
        {
            cp += n - tmp;
            tmp = cal(cp);
        }
        printf("%I64d
", cp);
    }
    return 0;
}