POJ 2891 Strange Way to Express Integers

2018-02-07 11:32:54来源:cnblogs.com作者:自为风月马前卒人点击

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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

28 711 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;}
   

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