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#include <iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
#include <math.h>
#include<string.h>
#include <algorithm>
#define INF 1000000000
using namespace std;
//typedef long long LL;
long long ggcd(long long a,long long b,long long&x,long long&y)
{
long long d,t;
if(b==0)
{
x=1;
y=0;
return a;
}
d=ggcd(b,a%b,x,y);
t=x-a/b*y;
x=y;
y=t;
return d;
}
int main()
{
long long a1,r1,a2,r2,a,b,c,x0,y0,m,t;
int i,n;
while(scanf("%d",&n)!=EOF)
{
int bj=1;
scanf("%lld%lld",&a1,&r1);
for(i=1;i<n;i++)
{
scanf("%lld%lld",&a2,&r2);
a=a1;
b=a2;
c=r2-r1;
m=ggcd(a,b,x0,y0);
if(c%m)
bj=0;
t=b/m;
x0*=(c/m);
x0=(x0%t+t)%t;
r1=a1*x0+r1;
a1=a1*(a2/m);
}
if(bj==0)
{
printf("-1\n");
}
else
printf("%lld\n",r1);
}
return 0;
}
Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:
Choose k different positive integers a1, a2, …, ak. For some non-negative m, divide it by every ai (1 ≤ i ≤ k) to find the remainder ri. If a1, a2, …, ak are properly chosen, m can be determined, then the pairs (ai, ri) can be used to express m.
“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”
Since Elina is new to programming, this problem is too difficult for her. Can you help her?
Input
The input contains multiple test cases. Each test cases consists of some lines.
Output
Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.
Sample Input
2 8 7 11 9
Sample Output
31
从后往前推有一个固定的式子；
x=1;y=0；
.
.
.
.
x=y;y=x-a/b*y;
Strange Way to Express Integers扩展欧几里德
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原文地址：http://www.cnblogs.com/luckylv/p/4842177.html
