hdu 5446 Unknown Treasure（lucas+中国剩余定理）

题目链接：hdu 5446 Unknown Treasure

lucas+中国剩余定理裸题，注意在中国剩余定理里面，有可能两数相乘爆long long，要用按位乘的方式，但是这样的话exgcd返回值如果是负数就会出错，所以乘之前要取模成正的。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5;

ll fac[maxn], inv[maxn];

ll pow_mod(ll a, int n, int mod) {
ll ret = 1;
while (n) {
if (n&1) ret = ret * a % mod;
a = a * a % mod;
n >>= 1;
}
return ret;
}

void init(int n) {
fac[0] = 1;
for (int i = 1; i < n; i++) fac[i] = fac[i-1] * i % n;
inv[n-1] = pow_mod(fac[n-1], n-2, n);
for (int i = n - 2; i >= 0; i--) inv[i] = inv[i+1] * (i+1) % n;
}

ll C (int n, int m, int mod) {
if (m > n || m < 0 || n < 0) return 0;
return fac
* inv[m] % mod * inv[n-m] % mod;
}

ll lucas(ll n, ll m, int mod) {
if (m == 0) return 1;
return lucas(n / mod, m / mod, mod) * C(n % mod, m % mod, mod) % mod;
}

ll exgcd(ll a, ll b, ll& x, ll& y) {
if (b == 0) { x = 1; y = 0; return a; }
ll d = exgcd(b, a % b, y, x);
y -= x * (a / b);
return d;
}

ll mul(ll a, ll b, ll mod) {
a = (a % mod + mod) % mod;
b = (b % mod + mod) % mod;

ll ret = 0;
while(b){
if(b&1){
ret += a;
if(ret >= mod) ret -= mod;
}
b >>= 1;
a <<= 1;
if(a >= mod) a -= mod;
}
return ret;
}

ll china(int n, ll* a, ll* m) {
ll M = 1, d, y, x = 0;
for (int i = 0; i < n; i++) M *= m[i];
for (int i = 0; i < n; i++) {
ll w = M / m[i];
exgcd(m[i], w, d, y);
x = (x + mul(mul(y, w, M), a[i], M));
}
return (x + M) % M;
}

int main () {
int cas, k;
ll n, m, a[15], p[15];
scanf("%d", &cas);
while (cas--) {
scanf("%lld%lld%d", &n, &m, &k);
for (int i = 0; i < k; i++) {
scanf("%lld", &p[i]);

init(p[i]);
a[i] = lucas(n, m, p[i]);
}

printf("%lld\n", china(k, a, p));
}
return 0;
}