20.11.13模拟 数对

给定一个n，求a+b=n,a+b ab 的正整数 数对的数量

设d=gcd(a,b),a=xd,b=yd,所以x+y xyd 又因为gcd(x,y)=1 ,所以x+y d

因为(x+y)d <=n 所以x+y<=(sqrt n)

所以我们可以枚举k=x+y,因为x，y互质，所以x与x+y互质，所以合法的(x,y)有(varphi(k))对，合法的d有(lfloor frac{n}{k^2} floor)

#include<iostream>
#include<cstdio>
#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i(j);i<=k;++i)
#define drp(i,j,k) for(register int i(j);i>=k;--i)
#define repg(x) for(register int i(G.head[x]);i;i=G.next[i])
#define bug cout<<"~~~~~~~~~~~~~"<<'
';
#define bugout(x) cout<<x<<'
';
using std::cin;
using std::cout;
typedef long long lxl;
template<typename T>
inline T  max( T a, T b) {
	return a > b ? a : b;
}
template<typename T>
inline T  min( T a, T b) {
	return a < b ? a : b;
}

inline char gt() {
	static char buf[1 << 21], *p1 = buf, *p2 = buf;
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
template <typename T>
inline void  read(T &x) {
	register char ch = gt();
	x = 0;
	int w(0);
	while(!(ch >= '0' && ch <= '9'))w |= ch == '-', ch = gt();
	while(ch >= '0' && ch <= '9')x = x * 10 + (ch & 15), ch = gt();
	w ? x = ~(x - 1) : x;
}
template <typename T>
inline void out(T x, char cc) {
	if(x < 0) x = -x, putchar('-');
	char ch[20];
	int num(0);
	while(x || !num) ch[++num] = x % 10 + '0', x /= 10;
	while(num) putchar(ch[num--]);
	putchar(cc);
}
const int N = 1e7 + 79;
lxl n;
int phi[N + 55], v[N], prime[N], cnt;
inline void init(int MX) {
	phi[1] = 1;
	rep(i, 2, MX) {
		if(!v[i]) {
			v[i] = i;
			prime[++cnt] = i;
			phi[i] = i - 1;
		}
		rep(j, 1, cnt) {
			if(prime[j] > v[i] || 1ll * prime[j]*i > 1ll * MX) break;
			v[i * prime[j]] = prime[j];
			phi[i * prime[j]] = phi[i] * (i % prime[j] ? prime[j] - 1 : prime[j]);
		}
	}
}

int main() {
	freopen("c.in", "r", stdin);
	freopen("c.out", "w", stdout);
	read(n);
	int t = sqrt(n);
	init(t + 5);
	lxl ans(0);
	rep(i, 2, t) {
		ans += phi[i] * (n / (1ll*i * i));
	}
	out(ans, '
');
	return 0;
}