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POJ--Strange Way to Express Integers

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Gxjun
发布2018-03-21 12:59:49
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发布2018-03-21 12:59:49
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文章被收录于专栏:ml

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, …, ak. For some non-negative m, divide it by every ai (1 ≤ ik) 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.

  • Line 1: Contains the integer k.
  • Lines 2 ~ k + 1: Each contains a pair of integers ai, ri (1 ≤ ik).

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

代码语言:javascript
复制
2
8 7
11 9

Sample Output

代码语言:javascript
复制
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:

代码语言:javascript
复制
 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:

代码语言:javascript
复制
 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 }

代码:

代码语言:javascript
复制
 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 }
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