zoukankan      html  css  js  c++  java
  • Codeforces 837 E Vasya's Function

    Discription

    Vasya is studying number theory. He has denoted a function f(a, b) such that:

    • f(a, 0) = 0;
    • f(a, b) = 1 + f(a, b - gcd(a, b)), where gcd(a, b) is the greatest common divisor of a and b.

    Vasya has two numbers x and y, and he wants to calculate f(x, y). He tried to do it by himself, but found out that calculating this function the way he wants to do that might take very long time. So he decided to ask you to implement a program that will calculate this function swiftly.

    Input

    The first line contains two integer numbers x and y (1 ≤ x, y ≤ 1012).

    Output

    Print f(x, y).

    Example
    Input
    3 5
    Output
    3
    Input
    6 3
    Output
    1


    水水数论,可以发现随着运算的过程,gcd单调不减,所以我们可以每次处理gcd相同的一段。
    #include<bits/stdc++.h>
    #define ll long long
    using namespace std;
    ll num[21],N=0,mn;
    ll a,b,d,ans=0,A,B;
    
    ll gcd(ll x,ll y){
    	return y?gcd(y,x%y):x;
    }
    
    inline void dvd(ll x){
    	for(int i=2;i*(ll)i<=x;i++) if(!(x%i)){
    		num[++N]=i;
    		while(!(x%i)) x/=i;
    		if(x==1) break;
    	}
    	if(x!=1) num[++N]=x;
    }
    
    inline void solve(){
    	dvd(a);
    	
    	d=gcd(a,b);
    	while(b){
    		A=a/d,B=b/d,mn=1ll<<60;
    		if(A==1){
    			ans+=B;
    			break;
    		}
    		
    		for(int i=1;i<=N;i++) if(!(A%num[i])) mn=min(mn,B%num[i]);
    		ans+=mn,b-=mn*d,d=gcd(b,a);
    	}
    	
    	printf("%I64d
    ",ans);
    }
    
    int main(){
    	scanf("%I64d%I64d",&a,&b);
    	solve();
    	return 0;
    }
    
     
  • 相关阅读:
    rockGenmel stone.txt
    WHICHDAY.txt
    WORKDAYS.txt
    WAIT_YN.txt
    WEEKDAYS.txt
    WHEREXY.txt
    KeySelected.txt
    WINDOW.txt
    UPPER.txt
    ParentShapes It.txt
  • 原文地址:https://www.cnblogs.com/JYYHH/p/8607727.html
Copyright © 2011-2022 走看看