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\n", cp);
}
return 0;
}