• 题解 P4450 【双亲数】


    [ans=sum_{i=1}^{a}sum_{j=1}^b[gcd(a,b)=d] ]


    我们假设

    [aleq b ]

    [egin{aligned} ans & = sum_{i=1}^{a}sum_{j=1}^b[gcd(a,b)=d] \ & = sum_{i=1}^{leftlfloordfrac{a}{d} ight floor}sum_{j=1}^{leftlfloordfrac{b}{d} ight floor}[gcd(i,j)=1] \ & = sum_{i=1}^{leftlfloordfrac{a}{d} ight floor}sum_{j=1}^{leftlfloordfrac{b}{d} ight floor}sum_{t|gcd(i,j)}mu(t) \ & = sum_{t=1}^{leftlfloordfrac{a}{d} ight floor}mu(t)leftlfloordfrac{a}{dt} ight floorleftlfloordfrac{b}{dt} ight floor end{aligned} ]

    数论分块

    #include<cstdio>
    #define Starseven main
    #define ll long long
    namespace lyt {
    	void read(int &x){
    	char ch=getchar();int re=0,op=1;
    	while(ch<'0'||ch>'9'){if(ch=='-') op=-1;ch=getchar();}
    	while(ch>='0'&&ch<='9'){re=(re<<3)+(re<<1)+ch-'0';ch=getchar();}
    	x = re * op;
    	return ;
    	}
    	void read(long long &x){
    	char ch=getchar();long long re=0,op=1;
    	while(ch<'0'||ch>'9'){if(ch=='-') op=-1;ch=getchar();}
    	while(ch>='0'&&ch<='9'){re=(re<<3ll)+(re<<1ll)+ch-'0';ch=getchar();}
    	x = re * op;
    	return ;
    	}
    	void write(int x){
    		if(x<0){putchar('-');x=-x;}
    		if(x>9) write(x/10);
    		putchar(x%10+'0');
    		return ;
    	}//记得自己加空格和换行 
    	void write(long long x){
    		if(x<0){putchar('-');x=-x;}
    		if(x>9) write(x/10);
    		putchar(x%10+'0');
    		return ;
    	}//记得自己加空格和换行
    	int max(int x,int y){return x<y?y:x;}
    	int min(int x,int y){return x<y?x:y;}
    	int abs(int x){return x<0?-x:x;}
    	long long max(long long x,long long y){return x<y?y:x;}
    	long long min(long long x,long long y){return x<y?x:y;}
    	long long abs(long long x){return x<0?-x:x;}
    	double abs(double x){return x<0?-x:x;}
    	void swap(int &a,int &b) {a ^= b ^= a ^= b;}
    	void swap(long long &a,long long &b) {a ^= b ^= a ^= b;}
    }using namespace lyt;
    const int maxn = 1e6;
    int a, b, d, prime[maxn + 20], num;
    ll mu[maxn + 20];
    bool vis[maxn + 20];
    
    void Init() {
    	mu[1] = 1;
    	for (int i = 2; i <= maxn; i++) {
    		if (!vis[i]) {
    			prime[++num] = i;
    			mu[i] = -1;
    		}
    		for (int j = 1; j <= num && prime[j] * i <= maxn; j++) {
    			int x = prime[j] * i;
    			vis[x] = true;
    			if (i % prime[j] == 0) {
    				mu[x] = 0;
    				break;
    			} 
    			mu[x] = -mu[i];
    		}
    	}
    	for (int i = 1; i <= maxn; i++) {
    		mu[i] += mu[i - 1];
    	}
    	return ;
    }
    
    int Starseven(void) {
    	read(a);
    	read(b);
    	read(d);
    	Init();
    	if (a > b) swap(a,b);
    	a /= d; b /= d;
    	ll ans = 0;
    	for (int i = 1, j; i <= a; i = j + 1) {
    		j = min(a / (a / i) , b / (b / i));
    		ll ta = (ll)a / i, tb = (ll)b / i;
    		ans += (mu[j] - mu[i - 1]) * ta * tb;
    	}
    	write(ans);
    	puts("");
    	return 0;	
    }
    
  • 相关阅读:
    windows7 dos修改mysql root密码
    Windows7 安装完成mysql dos启动服务出现服务名称无效
    windows 环境下.Net使用Redis缓存
    SQLSERVER 使用 ROLLUP 汇总数据,实现分组统计,合计,小计
    Web Form 和asp.net mvc 差别
    SQL Server如何编辑超过前200行的数据
    附加数据库失败,操作系统错误 5:"5(拒绝访问。)"的解决办法
    MVC 架构
    关于打印代码
    关于WCF一些基础。
  • 原文地址:https://www.cnblogs.com/starseven/p/13560247.html
Copyright © 2020-2023  润新知