Educational Codeforces Round 26 E

最新推荐文章于 2024-06-13 10:28:16 发布

hansy2015

最新推荐文章于 2024-06-13 10:28:16 发布

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文章标签： codeforces

本文链接：https://blog.youkuaiyun.com/gh6267/article/details/76710382

本文介绍了一种高效计算特定函数f(x, y)的方法，该函数递归地依赖于两个参数的最大公约数。通过去除参数间的共同质因数并计算这一过程中累积的步数，实现了快速求解。

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E. Vasya’s Function
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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).

Examples
input
3 5
output
3
input
6 3
output
1
我们知道a%b==0时计算停止，我们每计算一次就是消除a，b质因子之间的“差异”每消除一个差异就加一，所以我们只需计算出这种“差异”即可

#include <iostream>
#include <stdio.h>
#include <bits/stdc++.h>
#define int long long
using namespace std;
signed main(){
    int x,y;
    scanf("%I64d %I64d",&x,&y);
    int ans=0;
    while(y){
        int g=__gcd(x,y);
        x/=g;
        y/=g;
        int temp1=x,temp2=y;
        for(int i=2;i*i<=x;i++){
            if(temp1%i) continue;
            temp2=min(temp2,y%i);
            while((temp1%i)==0) temp1/=i;
        }
        if(temp1>1) temp2=min(temp2,y%temp1);
        y-=temp2;
        ans+=temp2;
    }
    cout<<ans<<endl;
    return 0;
}