POJ--Strange Way to Express Integers

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 8370		Accepted: 2508

Description

 

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, …, a_k. For some non-negative m, divide it by every a_i (1 ≤ i ≤ k) to find the remainder r_i. If a₁, a₂, …, a_k are properly chosen, m can be determined, then the pairs (a_i, r_i) 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.

• Line 1: Contains the integer k.
• Lines 2 ~ k + 1: Each contains a pair of integers a_i, r_i (1 ≤ i ≤ k).

 

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

Hint

All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

Source

POJ Monthly--2006.07.30, Static

同余问题：

关于欧几里得扩展的代码模板：

 1 void exgcd(int a,int b,int * x,int  *y,int  * d)

 2 {

 3    if(b==0)

 4 {

 5  x=1,y=0,d=a;

 6 }

 7 else

 8 {

 9    exgcd(b,a%b,x,y,d);

10 int temp=x;

11 x=y,y=temp-a/b*y;

12 }

13 }

解一元多项式时的代码：

 1 int sove()

 2 {

 3     scanf("%d%d",&a1,&r1);

 4     for(i=0;i<n;i++)

 5     {

 6         scanf("%d%d",&a2,r2);

 7         a=a1,b=a2,c=r2-r1;

 8         exgcd(a,b,x,y,d);

 9         if(c%d!=0)

10         {

11             ifhave=0;

12         }

13         int t=b/d;

14         x=(x*(c/d)%t+t)%t;

15         r1=r1+x*a1;

16         a1=a1*(a2/d);

17     }

18     if(!ifhave)

19     {

20         r1=-1;

21     }

22    return r1;  //即为解的个数

23 }

View Code

 

代码：

 1 #include<stdio.h>

 2 #define LL long long 

 3 LL x,y,q;

 4 void exgcd(LL a,LL b)

 5 {

 6     if(b==0)

 7     {

 8         x=1,y=0,q=a;

 9     }

10     else

11     {

12         exgcd(b,a%b);

13         LL temp=x;

14         x=y,y=temp-a/b*y;

15     }

16 }

17 

18 int main()

19 {

20     LL a1,r1,a2,r2,n,i;

21     bool ifhave;

22   while(scanf("%I64d",&n)!=EOF)

23   {

24   

25       scanf("%I64d%I64d",&a1,&r1);

26       ifhave=true;

27      for(i=1;i<n;i++)

28      {

29          scanf("%I64d%I64d",&a2,&r2);

30          exgcd(a1,a2); 

31          if((r2-r1)%q)

32          {

33              ifhave=false;

34          }

35          LL t=a2/q;

36          x=(x*((r2-r1)/q)%t+t)%t;

37          r1+=a1*x;

38          a1*=(a2/q);

39      }

40      if(!ifhave) r1=-1;

41      printf("%I64d\n",r1);

42   }

43 return 0;

44 }

View Code