42 lines
895 B
C++
42 lines
895 B
C++
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// https://vjudge.csgrandeur.cn/problem/HDU-1576
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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/*
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测试用例
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2
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1000 53
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87 123456789
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答案
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7922
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6060
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总结：
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1、如果使用费马小定理+快速幂求逆元，一定要注意使用LL，防止爆int。
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*/
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const int MOD = 9973;
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LL qmi(LL a, LL k, LL p) {
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LL res = 1;
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while (k) {
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if (k & 1) res = res * a % p; //如果res是int,可能会越界
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k >>= 1;
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a = a * a % p; //如果a是int, 可能会越界
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}
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return res;
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}
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int main() {
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int T;
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cin >> T;
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while (T--) {
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LL a, b;
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cin >> a >> b;
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LL x;
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x = qmi(b, MOD - 2, MOD); // b理解为分母,同余MOD的逆元,变除法为乘法
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LL res = a * x % MOD; // a* b在模MOD下的逆元 x
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printf("%lld\n", res);
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}
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return 0;
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}
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