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  • Bzoj2440: [中山市选2011]完全平方数

    题面戳我

    solution

    考虑二分答案,二分了一个答案n
    在n以内,我们需要快速求出题目要我们求的数的个数

    所以可以用莫比乌斯反演
    设f[i]表示只含有i^2不含其它完全平方数的数的个数
    (g[i] = sum_{i}^{i|d} f[d])
    则g[i]表示所有i^2的倍数(g[i]=lfloor frac{n}{i^2} floor)
    (f[i] = sum_{i}^{i|d} mu[d]f[d])
    f[1]即为二分所求

    # include <bits/stdc++.h>
    # define RG register
    # define IL inline
    # define Fill(a, b) memset(a, b, sizeof(a))
    using namespace std;
    typedef long long ll;
    const int _(1e4 + 10), __(1e6 + 10);
    
    IL ll Read(){
    	char c = '%'; ll x = 0, z = 1;
    	for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
    	for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
    	return x * z;
    }
    
    int k, prime[__], isprime[__], num, mu[__];
    ll ans;
    
    IL void Prepare(){
    	mu[1] = 1;
    	for(RG int i = 2; i <= __; ++i){
    		if(!isprime[i]) prime[++num] = i, mu[i] = -1;
    		for(RG int j = 1; j <= num && i * prime[j] <= __; ++j){
    			isprime[i * prime[j]] = 1;
    			if(!(i % prime[j])){  mu[i * prime[j]] = 0; break;  }
    			else mu[i * prime[j]] = -mu[i];
    		}
    	}
    }
    
    IL ll Check(RG ll n){
    	RG ll f = 0, g = 0;
    	for(RG ll i = 1; i * i <= n; ++i) g = n / (i * i), f += mu[i] * g;
    	return f;
    }
    
    int main(RG int argc, RG char *argv[]){
    	Prepare();
    	for(RG int T = Read(), i = 1; i <= T; ++i){
    		k = Read(); ans = 0;
    		RG ll l = 1, r = 1e10;
    		while(l <= r){
    			RG ll mid = (l + r) >> 1;
    			if(Check(mid) >= k) ans = mid, r = mid - 1;
    			else l = mid + 1;
    		}
    		printf("%lld
    ", ans);
    	}
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8245166.html
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