TOJ 2644

题目连接:

http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=2644

题目类型:

数论 - 扩展欧几里得

数据结构:

无

思路分析:

将题目抽象成 公式:

( x + t * m ) % L = ( y + t * n ) % L

=>  ( x + m * t ) - ( y + n * t ) = p * L  

=>  ( n - m ) * t + p * L = ( x - y )

设 n - m = A, x - y = B

求满足 A * t + p * L = B 的最小解

即求同余式 A * t = B ( mol L )  的最小正整数解

具体过程:

1. ( n - m ) * t + p * L = ( x - y )  的解 ( 利用扩展欧几里得 )

2. 若 ( x - y ) % gcd( n - m, L ) == 0, 则有解

3. 得到解后, 令 M = gcd( n - m, L ) , X = X * ( x - y ) / M;

4. ( X % ( L / M )  + L / M ) % ( L / M )  即为最后解

证明:

暂无

源代码:

#include 
#include 

using namespace std;

__int64 _extend_gcd( __int64 a, __int64 b, __int64 * x, __int64 * y ) 
{
	__int64 x0,x1,x2,y0,y1,y2;
	__int64 r0,r1,r2,q;
	
	if((a==0)&&(b==0)) { *x=0;*y=0; return -1; }
	if((a==0)&&(b!=0)) { *x=0;*y=1; return b; }
	if((a!=0)&&(b==0)) { *x=1;*y=0; return a; }
	if((a!=0)&&(b!=0))
	{
		x0=0;x1=1;r0=a; y0=1;r1=b; r2=r0%r1; y1=0-r0/r1; x2=1;y2=y1;
		if(r2==0)  { x=0; *y=1; return r1; }
		while((r1%r2)!=0)
		{
			r0=r1;r1=r2; q=r0/r1;
			x2=x0-x1*q; y2=y0-y1*q;
			x0=x1;x1=x2; y0=y1;y1=y2;
			r2=r0%r1;
		}
		*x=x2; *y=y2;
		return r2;
	}
}

int main()
{
	__int64 a, b;
	__int64 x, y, m, n, L;
	
	while( scanf( "%I64d%I64d%I64d%I64d%I64d", &x, &y, &m, &n, &L ) != EOF )
	{
		__int64 c = _extend_gcd( n - m, L, &a, &b );
		
		if( ( x - y ) % c || m == n )
		{
			puts( "Impossible" );
		}
		else
		{
			__int64 s = L / c;
			a = a * ( ( x - y ) / c );
			a = ( a % s + s ) % s;
			printf( "%I64d\n", a );
		}
	}
	
	return 0;
}