SGU 140. Integer Sequences 线性同余,数论 难度:2

140. Integer Sequences

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB

 

A sequence A is called an integer sequence of length N if all its elements A₁ A₂ .. A_N are non-negative integers less than 2 000 000 000. Consider two integer sequences of length N, A and X. The result of their multiplication (A*X) is an integer number R=A₁*X₁ + A₂*X₂ + .. + A_N*X_N. Your task is to solve the equation A*X=B (mod P), given the integer sequence A and the integer numbers B and P.

 

Input

The first line contains the integer numbers N (1<=N<=100) - the length of the integer sequences - P (1<=P<=10 000) and B (0<=B<=P-1). The second line contains the elements of the sequence A, separated by blanks: A₁ A₂ .. A_N.

 

Output

You should print one line containing the word "YES" if there exists at least one integer sequence X which is a solution to the equation, or print "NO" otherwise. If the answer is "YES", the next line should contain N non-negative integers separated by blanks: X₁ X₂ .. X_N.

 

Sample Input #1

2 7 4

7 3

Sample Output #1

YES

0 6

Sample Input #2

3 10 1

2 4 6

Sample Output #2

NO
线性同余方程,不断使前k个项余p得到最大公约数,同除去最大公约数,逆推即可

#include <cstdio>

using namespace std;

int extgcd(int a,int b,int& x,int& y){

	if(a==0){

		x=0;y=1;

		return b;

	}

	int t=extgcd(b%a,a,x,y);

	int temp=x;

	x=y-b/a*x;

	y=temp;

	return t;

}

int num[110],x[110],y[110];

int main(){

    int cgcd,n,p,b;

    scanf("%d%d%d",&n,&p,&b);

    for(int i=0;i<n;i++){

        scanf("%d",num+i);

        num[i]%=p;

    }

    cgcd=num[0];

    for(int i=1;i<n;i++){

        cgcd=extgcd(cgcd,num[i],x[i],y[i]);

    }

    cgcd=extgcd(cgcd,p,x[n],y[n]);

    if(b%cgcd!=0)puts("NO");

    else {

        puts("YES");

        b/=cgcd;

        y[0]=1;

        for(int i=n-1;i>=0;i--){

            while(x[i+1]<0)x[i+1]+=p;

            b*=x[i+1];

            b%=p;

            while(y[i]<0)y[i]+=p;

            y[i]=y[i]*b%p;

        }

        for(int i=0;i<n;i++){

            printf("%d%c",y[i],i==n-1?'\n':' ');

        }

    }

    return 0;

}