POJ 2891 Strange Way to Express Integers

2018-06-17 21:16:58来源:未知 阅读 ()

新老客户大回馈,云服务器低至5折

Strange Way to Express Integers
Time Limit: 1000MS   Memory Limit: 131072K
Total Submissions: 17963   Accepted: 6050

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 a1a2…, ak. For some non-negative m, divide it by every ai (1 ≤ i ≤ k) to find the remainder ri. If a1a2, …, ak are properly chosen, m can be determined, then the pairs (airi) 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 airi (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
 
题意
给出$a_i,r_i$
求$x\equiv r_{i}\left( mod\ a_{i}\right)$
其中$a_i$不互质
 
 
扩展CRT的应用,算是裸题吧
第一次一遍写对扩欧好感动啊。。。
 
#include<iostream>
#include<cstdio>
#define LL long long 
using namespace std;
const LL MAXN=1e6+10;
LL K,C[MAXN],M[MAXN],x,y;
LL gcd(LL a,LL b)
{
    return b==0?a:gcd(b,a%b);
}
LL exgcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0){x=1,y=0;return a;}
    LL r=exgcd(b,a%b,x,y),tmp;
    tmp=x;x=y;y=tmp-(a/b)*y;
    return r;
}
LL inv(LL a,LL b)
{
    LL r=exgcd(a,b,x,y);
    while(x<0) x+=b;
    return x;
}
int main()
{
    #ifdef WIN32
    freopen("a.in","r",stdin);
    #else
    #endif
    while(~scanf("%lld",&K))
    {
        for(LL i=1;i<=K;i++) scanf("%lld%lld",&M[i],&C[i]);
        bool flag=1;
        for(LL i=2;i<=K;i++)
        {
            LL M1=M[i-1],M2=M[i],C2=C[i],C1=C[i-1],T=gcd(M1,M2);
            if((C2-C1)%T!=0) {flag=0;break;}
            M[i]=(M1*M2)/T;
            C[i]= ( inv( M1/T , M2/T ) * (C2-C1)/T ) % (M2/T) * M1 + C1;
            C[i]=(C[i]%M[i]+M[i])%M[i];
        }
        printf("%lld\n",flag?C[K]:-1);
    }
    return 0;
}

 

 
 
 

标签:

版权申明:本站文章部分自网络,如有侵权,请联系:west999com@outlook.com
特别注意:本站所有转载文章言论不代表本站观点,本站所提供的摄影照片,插画,设计作品,如需使用,请与原作者联系,版权归原作者所有

上一篇:C++11新特性学习

下一篇:洛谷 P1430 序列取数