Submission #3226187

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/*
结论题Orz

设$f(x, y)$表示$(x, y)$辗转相除需要的步数，$fib(i)$表示第$i$个斐波那契数

常识：$f(fib[i], fib[i+1]) = i$。

定义一个数对是“好的”，当且仅当对于$(x, y)$，不存在更小的$x', y'$使得$f(x', y') > f(x, y)$

显然我们需要统计的数对一定是好的数对

定义一个数对是“优秀的”，当且仅当对于$(x, y)$，若$f(x, y) = k$, 满足$x, y \leqslant fib[k+2] + fib[k-1]$

结论！：一个好的数对辗转相除一次后一定是优秀的数对！

证明可以用反证法，也就是我先假设一个$f(a, b) = i$是好的，但是得到的数对$(x, y)$满足$y > fib[k+2] + fib[k-1]$

但是这样我们会得到一个$x' = f[i+2], y' = f[i+2]$满足$f(x', y')>f(a, b)$，所以不成立

那么现在要做的就是求“优秀的”数对的个数。

考虑直接用欧几里得算法的定义递推即可

*/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#define Pair pair<LL, LL> 
#define MP(x, y) make_pair(x, y)
#define fi first 
#define se second 
#define LL long long 
#define int long long 
using namespace std;
const int MAXN = 1e6 + 10, B = 90, mod = 1e9 + 7;
inline LL read() {
	char c = getchar(); LL x = 0, f = 1;
	while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}
vector<Pair> v[B + 1];
LL f[B + 1];
void Pre() {
	f[0] = f[1] = 1;
	for(int i = 2; i <= B; i++) f[i] = f[i - 1] + f[i - 2];
	v[1].push_back(MP(1, 2)); v[1].push_back(MP(1, 3)); v[1].push_back(MP(1, 4));
	for(int i = 1; i <= B - 3; i++) {
		for(int j = 0; j < v[i].size(); j++) {
			LL x = v[i][j].fi, y = v[i][j].se;
			LL tmp = x; x = y; y = tmp + y;
			while(y <= f[i + 3] + f[i - 1]) v[i + 1].push_back(MP(x, y)), y += x;
		}
	}
}
main() {
	// freopen("1.in", "r", stdin);
	Pre();
	int Q = read();
	while(Q--) {
		LL x = read(), y = read(), K;
		if(x > y) swap(x, y);
		for(K = 1; f[K + 1] <= x && f[K + 2] <= y; K++);
		cout << K << " ";
		if(K == 1) {cout << x * y % mod << endl; continue;}
		LL ans = 0;
		for(int i = 0; i < v[K - 1].size(); i++) {
			LL a = v[K - 1][i].fi, b = v[K - 1][i].se;
		//	printf("%I64d %I64d\n", a, b);
			if(b <= x) ans += (y - a) / b % mod;
			if(b <= y) ans += (x - a) / b % mod;
			//if(a + b <= x && b <= y) ans++;
			//if(a + b <= y && a <= x) ans++;
			ans %= mod;
		}
		cout << ans % mod<< endl;
	}
	return 0;
}
/*
1
12 11
*/

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