程式程式碼：

/****************************************************/
/* File        : Hiho_Week_97                       */
/* Author      : Zhang Yufei                        */
/* Date        : 2016-05-09                         */
/* Description : HihoCoder ACM program. (submit:g++)*/
/****************************************************/

#include<stdio.h>
#include<stdlib.h>

/*
 * Define the structure to record the result.
 * Parameters:
 *		@x & @y: The result (x, y).
 */
typedef struct node {
	long long x;
	long long y;
} result;

/*
 * This function computes the greatest common divisor of 2 numbers a and b.
 * Parameters:
 *		@a & @b: The 2 numbers to compute.
 * Returns:
 *		The greatest common divisor of a and b.
 */
long long gcd (long long a, long long b) {
	if(a % b == 0) {
		return b;
	} else {
		return gcd (b, a % b);
	}
}

/*
 * This function computes the lowest common multiple of 2 numbers a and b.
 * Parameters:
 *		@a & @b: The 2 numbers to compute.
 * Returns:
 *		The lowest common multiple of a and b.
 */
long long lcm (long long a, long long b) {
	long long g = gcd (a, b);
	return (a / g) * b;
}

/*
 * This function calculates the result of the equation:
 *	Ax + By = 1 using extend euclidean theory.
 * Parameters:
 * 		@A & @B: The parameter A, B in the equation.
 * Returns:
 * 		One result of this equation, if the equation don't have 
 *		a legal result, returns NULL.
 */ 
result* extend_euclidean (long long A, long long B) {
	result *r; 
	if(A % B == 0) {
		r = (result*) malloc (sizeof(result));
		r->x = 0;
		r->y = 1;
	} else {
		r = extend_euclidean (B, A % B);
		long long x = r->x;
		long long y = r->y;
		r->x = y;
		r->y = x - A / B * y;
	}
	return r;
}

/*
 * The main program.
 */
int main (void) {
	int N;
	scanf ("%d", &N);
	
	long long m1, r1, m2, r2;
	long long A, B, C;
	scanf ("%lld %lld", &m1, &r1);
	
	result *r;
	
	long long answer = 0;
	
	for(int i = 1; i < N; i++) {
		scanf ("%lld %lld", &m2, &r2);
		if(answer == -1) {
			continue;
		}
		A = m1;
		B = m2;
		C = r2 - r1;
		
		long long g = gcd (A, B);
		if(C % g != 0) {
			answer = -1;
			continue;
		}
		
		long long lc = lcm (A, B);
						
		A /= g;
		B /= g;
		C /= g;
		
		r = extend_euclidean (A, B);
		r->x *= C;
		r->x %= B;
		
		answer = m1 * r->x + r1;

		while(answer < 0) {
			answer += lc;
		}
		
		while(answer > lc) {
			answer -= lc;
		}
		
		m1 = lc;
		r1 = answer;
	} 
	
	printf("%lld\n", answer);
	
	return 0;
}