poj2891(一元线性同于方程组)

Strange Way to Express Integers

Time Limit: 1000MS Memory Limit: 131072KTotal Submissions: 8077 Accepted: 2398

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 everya_i (1 ≤ i ≤ k) to find the remainder r_i. Ifa₁, 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 findm 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

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.

解一元线性方程组：

x(mod a1)=r1(mod a1)

x(mod a2)=r2(mod a2)

.....

xn(mod an)=rn (mod an);

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long ll;void exgcd(ll a,ll b,ll &d,ll &x,ll &y){    if(!b){d=a;x=1;y=0;}    else    {        exgcd(b,a%b,d,y,x);        y-=x*(a/b);    }}int main(){    int n;    ll a1,a2,r1,r2,m;    while(scanf("%lld",&n)!=EOF)    {        ll a,b,c,d,x0,y0;        bool ifhave=true;        scanf("%lld%lld",&a1,&r1);        for(int i=1;i<n;i++)        {            scanf("%lld%lld",&a2,&r2);            a=a1,b=a2,c=r2-r1;            exgcd(a,b,d,x0,y0);            if(c%d!=0)            {                ifhave=false;            }            ll t=b/d;            x0=(x0*(c/d)%t+t)%t;            r1=a1*x0+r1;            a1=a1*(a2/d);        }        if(!ifhave) printf("-1\n");        else        {            m=r1;            printf("%lld\n",m);        }    }    return 0;}