sgu 119 Magic Pairs

最新推荐文章于 2021-08-05 15:28:46 发布

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最新推荐文章于 2021-08-05 15:28:46 发布

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本文解决了一个数学竞赛中关于整数对的问题，通过数学原理找到所有满足条件的整数对，并提供了算法实现。适用于数学竞赛准备和算法学习。

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题目描述：

119. Magic Pairs

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

“Prove that for any integer X and Y if 5X+4Y is divided by 23 than 3X+7Y is divided by 23 too.” The task is from city Olympiad in mathematics in Saratov, Russia for schoolchildren of 8-th form. 2001-2002 year.

For given N and pair (A₀, B₀) find all pairs (A, B) such that for any integer X and Y if A₀X+B₀Y is divided by N then AX+BY is divided by N too (0<=A,B<N).

Input

Each input consists of positive integer numbers N, A₀ and B₀ (N,A₀,B₀£10000) separated by whitespaces.

Output

Write number of pairs (A, B) to the first line of output. Write each pair on a single line in order of non-descreasing A (and B in case of equal A). Separate numbers by single space.

Sample Input

3
1 2

Sample Output

Author	: Michael R. Mirzayanov
Resource	: PhTL #1 Training Contests
Date	: Fall 2001

思路：其实这道题是参考大牛博客才搞出来的，果然自己数学还是不行啊。

结论就是所有a*i%n,b*i%n 都是解。

知道结论去证明就不难了。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<algorithm>
#include<vector>
#include<cstdlib>

#define inf 0xfffffff
#define CLR(a,b) memset((a),(b),sizeof((a)))

using namespace std;
int const nMax = 1000;
int const base = 10;
typedef int LL;
typedef pair<LL,LL> pij;

set<pij> ans;
int a,b,n;

int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}

int main(){
    cin>>n>>a>>b;
    ans.clear();
    int d=n/gcd(gcd(a,b),n);
    for(int i=0;i<d;i++){
        ans.insert(pij((a*i)%n,(b*i)%n));
    }
    set<pij>::iterator it;
    printf("%d\n",ans.size());
    for(it=ans.begin();it!=ans.end();it++){
        printf("%d %d\n",(*it).first,(*it).second);
    }
    return 0;
}