中国剩余定理
输入a,m 第i个方程表示为x ≡ ai(mod mi),求x。
两两互质:
const int maxn = 20;
ll a[maxn], m[maxn], n;
ll CRT(ll a[], ll m[], int n) {
ll M = 1;
for (int i = 0; i < n; i++) M *= m[i];
ll ret = 0;
for (int i = 0; i < n; i++) {
ll x, y;
ll tm = M / m[i];
extend_gcd(tm, m[i], x, y);
ret = (ret + tm * x * a[i]) % M;
}
return (ret + M) % M;
}
POJ:2891 两两不互质
当两两不互质时,不能够用上述方法求,但可以通过将两个......
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