UVA 10673 扩展欧几里得

题意：给出x 和k，求解p和q使得等式x = p[x / k] + q [ x / k]， 两个[x / k]分别为向下取整和向上取整

题解：扩展欧几里得

//meek///#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <sstream>
#include <vector>
using namespace std ;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair
typedef long long ll;

const int N = 110;
const int inf = 99999999;
const int mod= 1000000007;



ll ExpGcd(ll a,ll b,ll &x,ll &y)
{
    ll temp,p;
    if(b==0)
    {
        x=1; y=0;
        return a;
    }
    p=ExpGcd(b,a%b,x,y);
    temp=x; x=y; y=temp-(a/b)*y;
    return p;
}

int main() {
    int T;
    scanf("%d",&T);
    while(T--) {
        ll x,k,X,Y;
        scanf("%lld%lld",&x,&k);
        ll a = floor(1.0*x/k);
        ll b = ceil(1.0*x/k);
        ll d = ExpGcd(a,b,X,Y);
        //cout<<X<<" "<<Y<<endl;
        //if(X<0) X=(X%b+b)%b;
        X    *= x/d;
        Y    *= x/d;
        printf("%lld %lld\n",X,Y);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/zxhl/p/5074003.html