HDU 5446 Unknown Treasure Lucas定理+中国剩余定理

本文主要是介绍HDU 5446 Unknown Treasure Lucas定理+中国剩余定理，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

数学本来就弱，现在大半年不练，果断简单题都不会了。。。先用lucas求出x mod 每一个素数结果，然后再用中国剩余定理解出x，不过要自己写乘法，防溢出。。。

#include <cstdio>
#include <cstring>
using namespace std;
typedef __int64 LL;
LL a[12], mm[12];
LL mul(LL a, LL b, LL c)
{LL ans = 0;while(b){if(b&1){ans += a;ans %= c;}a += a;a %= c;b >>= 1;}return ans;
}
LL pow(LL a, LL b, LL c)
{LL ans = 1;while(b){if(b&1)ans = mul(ans, a, c);b >>= 1;a = mul(a, a, c);}return ans;
}LL cm(LL n, LL m, LL p)
{if(n < m)return 0;if(m > n-m)m = n-m;LL s1 = 1, s2 = 1;for(LL i = 0;i < m; i++){s1 = mul(s1, (n-i), p);s2 = mul(s2, (i+1), p);}return mul(s1, pow(s2, p-2, p), p);
}
LL lucas(LL n, LL m, LL p)
{if(m == 0)return 1;//puts("asf");return mul(cm(n%p, m%p, p), lucas(n/p, m/p, p), p);
}
void gcd(LL a, LL b, LL& d, LL& x, LL& y)
{if(!b){d = a;x = 1;y = 0;}else{gcd(b, a%b, d, y, x);y -= x * (a/b);}
}LL china(LL n, LL* a, LL* mm)
{LL M = 1, d, y, x = 0;for(int i = 0; i < n; i++)M *= a[i];for(int i = 0; i < n; i++){LL w = M /a[i];gcd(a[i], w, d, d, y);x = (x + mul(mul(y, w, M), mm[i], M)) % M;}return (x+M)%M;
}
int main()
{int T;scanf("%d", &T);while(T--){LL n, m, k;scanf("%I64d %I64d %I64d", &n, &m, &k);for(int i = 0; i < k; i++){scanf("%I64d", &a[i]);}for(int i = 0; i < k; i++){mm[i] = lucas(n, m, a[i]);}printf("%I64d\n", china(k, a, mm));}
}

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