HDU

本来是个求逆元的题，

但是这里可以 枚举 （A / B），已知 A % m = n ,

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <cmath>#include <set>#include <map>#include <stack>#include <queue>#include <ctype.h>#include <vector>#include <algorithm>#include <sstream>#define PI acos(-1.0)#define in freopen("in.txt", "r", stdin)#define out freopen("out.txt", "w", stdout)using namespace std;typedef long long ll;const int maxn = 500 + 7, INF = 0x3f3f3f3f, mod = 1e9 + 7;int T;ll b, t, n;int main() {    scanf("%d", &T);    while(T--) {        scanf("%lld%lld", &n, &b);        for(ll i = 1; i <= 9973; ++i) {            ll t = i*b-n;            if(t % 9973 == 0) {                printf("%lld\n", (i%9973)); break;            }        }    }    return 0;}