- 前两个移动很像辗转相除法（这个套路在 Codeforces 上已经出烂了）<br>
- 后两个移动可以让 g 乘上 $2^k$
class Solution {
public:
bool isReachable(int X, int Y) {
while(!(X&1))X >>= 1;
while(!(Y&1))Y >>= 1;
return gcd(X,Y) == 1;
}
};
- https://codeforces.com/contest/1766/problem/D
题意:
找到连续的最长gcd(a+k，b+k) == 1(a < b,k = {0,1,2,...})
- 思路：
- gcd(a+k,b+k) == gcd(a+k,b - a)
- a - b = 1时特判
- 可以推出$gcd(a+k,b+k) == gcd(a+k,b - a)$,具体证明见https://codeforces.com/blog/entry/110066
- 设两个的结果分别为g和h，显然a+k和b+k都是g的倍数，那么让其中一个大的倍数减掉一个小的倍数，显然还是g的倍数，充分性成立
- 反证法也同理，a+k和b-a肯定都是h的倍数，一个倍数加上另一个倍数，肯定还是h的倍数，必要性成立
- 接下来只需要找到b-a的质因子判断a+k是否有重复因子
- 一个一个加显然超时，我们可以先预处理出每个数字的最大质因子，然后利用传递性，得到关于b-a的所有质因子
- 设当前质因子为temp，更新答案的方法就是$ans = min(ans,(temp - a%temp) % temp)$
#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<" "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<" "<<#d<<" = "<<d<<" "<<#e<<" = "<<e<<endl;
#define debug0(x) cout << "debug0: " << x << endl
#define fr(t, i, n)for (long long i = t; i < n; i++)
#define YES cout<<"Yes"<<endl
#define nO cout<<"no"<<endl
#define fi first
#define se second
//#define int long long
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)
const int N = 2e5+10,M = 1e7+10;
int to[M];
void solve()
{
int x,y;scanf("%d%d",&x,&y);
if(y == x+1)
{
printf("-1\n");
return ;
}
int now = y - x;
int ans = 1e9;
for(;now != 1;now /= to[now])
ans = min(ans, (now - x % to[now]) % to[now]);
printf("%d\n",ans);
}
signed main()
{
/*
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
*/
to[1] = 1;//每个数字到它最大的质因子
for(int i = 2;i < M;i ++)if(!to[i])
{
for(int j = i;j < M;j += i)to[j] = i;
}
int T = 1;
scanf("%d",&T);
//cin >> T;
while(T--){
solve();
}
return 0;
}
- https://codeforces.com/contest/1762/problem/D
题意：
给一个长度为n的permutation，每次一个询问，得到结果为gcd（i，j），请在2*n次之内找到那个是0（或者哪两个之中之一是0）思路：
- 三个指针i，j，k（i<j<k）
- 令x=gcd(a[i],a[j]),y=gcd(a[i],a[k]);
- 这样就可以写出一个答案了
#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<" "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<" "<<#b<<" = "<<b<<" "<<#c<<" = "<<c<<" "<<#d<<" = "<<d<<" "<<#e<<" = "<<e<<endl;
#define debug0(x) cout << "debug0: " << x << endl
#define fr(t, i, n)for (long long i = t; i < n; i++)
#define YES cout<<"Yes"<<endl
#define nO cout<<"no"<<endl
#define fi first
#define se second
// #define int long long
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)
const int N = 2e5+10,mod = 998244353;
bool st[N];
int ask(int a, int b) {
cout << "? " << a << " " << b << endl;
int ans = 0;
cin >> ans;
return ans;
}
void solve()
{
memset(st,0,sizeof st);
int n;cin >> n;
int a[3] = {1,2,3};
for(;;)
{
sort(a,a+3);
if(a[2] > n)break;
int x = ask(a[0],a[1]),y = ask(a[0],a[2]);
if(x == y)
{
st[a[0]] = 1;
a[0] = a[2] + 1;
}else if(x < y)
{
st[a[1]] = 1;
a[1] = a[2] + 1;
}else{
st[a[2]] = 1;
a[2] = a[2] + 1;
}
}
cout << "! " << a[0] << " " << a[1] << endl;
int t;cin >> t;
}
signed main()
{
/*
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
*/
int T = 1;cin >> T;
while(T--){
solve();
}
return 0;
}
标签：GCD,套路,cout,相除,st,int,ans,now,gcd From： https://www.cnblogs.com/cfddfc/p/17064646.html