快速切题 sgu119. Magic Pairs

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

3 
0 0
1 2
2 1

//started 22:24
//read error 1 23:15
//read answer 0:41 晕,既然a0x+b0y|n,ax+by|k*n,直接乘上k再modn即可....
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int extgcd(int a,int b,int &x,int &y){
    int d=a;
    if(b!=0){
        d=extgcd(b,a%b,y,x);
        y-=(a/b)*x;
    }
    else {
        x=1;y=0;
    }
    return d;
}
typedef pair<int ,int> P;
P heap[10001];
int cnt;
int main(){
    int n,a0,b0;
    scanf("%d%d%d",&n,&a0,&b0);
    int ty,tx,tmp1,tmp2;
    int abgcd=extgcd(a0,b0,tx,ty);
    int nabgcd=extgcd(abgcd,n,tmp1,tmp2);
    n=n/nabgcd;
    a0=a0/nabgcd;
    b0=b0/nabgcd;
    for(int i=0;i<n;i++){
        int a=a0*i%n;
        int b=b0*i%n;
        heap[i]=P(a,b);
    }
    sort(heap,heap+n);
    printf("%d\n",n);//必然有n个解
    for(int i=0;i<n;i++){
        printf("%d %d\n",heap[i].first*nabgcd,heap[i].second*nabgcd);
    }
    return 0;
}