Candy Distribution

Kids like candies, so much that they start beating each other if the candies are not fairly distributed. So on your next party, you better start thinking before you buy the candies.

If there are KK kids, we of course need K⋅XK⋅X candies for a fair distribution, where XX is a positive natural number. But we learned that always at least one kid looses one candy, so better be prepared with exactly one spare candy, resulting in (K⋅X)+1(K⋅X)+1 candies.

Usually, the candies are packed into bags with a fixed number of candies CC. We will buy some of these bags so that the above constraints are fulfilled.

Input

The first line gives the number of test cases tt (0<t<1000<t<100). Each test case is specified by two integers KK and CC on a single line, where KK is the number of kids and CC the number of candies in one bag (1≤K,C≤1091≤K,C≤109). As you money is limited, you will never buy more than 109109 candy bags.

Output

For each test case, print one line. If there is no such number of candy bugs to fulfill the above constraints, print “IMPOSSIBLE” instead. Otherwise print the number of candy bags, you want to buy. If there is more than one solution, any will do.

Sample Input 1	Sample Output 1
5 10 5 10 7 1337 23 123454321 42 999999937 142857133	IMPOSSIBLE 3 872 14696943 166666655

 

题解：就是求一个不定方程 k*x-c*y=1  有几个注意的地方 对c==1的情况进行特判  不能有puts输出 还有就是不能有(x%d+d)%d来更新x

#include<iostream>
#include<cstring>
#include<cstdio> 
using namespace std;
typedef long long ll;
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);
    ll t=y;
    y=x-(a/b)*y;
    x=t;
    return r;
}
int main(){
    ios::sync_with_stdio(false);
    ll t;
    cin>>t;
    while(t--){
        ll k,c;
        cin>>k>>c;
        ll x,y;
        ll r=exgcd(c,k,x,y);
        if(r!=1){
//            puts("IMPOSSIBLE");
            cout<<"IMPOSSIBLE"<<'\n';
        }
        else {
            if(c==1) {
                cout<<k+1<<'\n';
            }
            else{
                while(x<=0) x+=k;
                cout<<x<<'\n';
            }
        }
    }
    return 0; 
}