HDU 5446 Unknown Treasure （2015年长春赛区网络赛J题）

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1.题目描述：点击打开链接

2.解题思路：本题利用Lucas+CRT+移位乘法解决。然而。。。比赛时候意识到了可能会爆long long，但是还没学移位乘法，受到了100点伤害==

3.代码：

#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<algorithm>#include<cassert>#include<string>#include<sstream>#include<set>#include<bitset>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<cctype>#include<complex>#include<functional>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define me(s)  memset(s,0,sizeof(s))#define rep(i,n) for(int i=0;i<(n);i++)typedef long long ll;typedef unsigned int uint;typedef unsigned long long ull;typedef pair <int, int> P;ll n, m;int k;int primes[12];int a[12];ll pow_mod(ll a, ll n, ll p){ll res = 1;while (n>0){if (n & 1)res = res*a%p;a = a*a%p;n >>= 1;}return res;}void gcd(ll a, ll b, ll&d, ll&x, ll&y){if (!b) { x = 1, y = 0, d = a; }else{gcd(b, a%b, d, y, x);y -= x*(a / b);}}ll C(ll n, ll m, int p){if (m>n)return 0;ll t1 = 1, t2 = 1;for (int i = n; i>m; i--){t1 = (t1*i) % p;t2 = (t2*(i - m)) % p;}return (t1%p)*(pow_mod(t2, p - 2, p) % p) % p;}ll Lucas(ll n, ll m, int p){if (!m)return 1;ll ans = C(n%p, m%p, p);return ans*Lucas(n / p, m / p, p) % p;}ll M;ll mul(ll a, ll b, ll p){ll res = 0;for (; a; a >>= 1){if (a & 1){res += b;if (res >= p)res -= p;}b <<= 1;if (b >= p)b -= p;}return res;}ll China(int n, int*a, int*m){M = 1;for (int i = 0; i<n; i++)M *= m[i];ll x = 0, d, y;for (int i = 0; i<n; i++){ll w = M / m[i];gcd((ll)m[i], w, d, d, y);int sign;if (y < 0)sign = -1;else sign = 1;x = (x + mul(y*sign, w*a[i] % M, M)*sign) % M;}return (x + M) % M;}int main(){int T;scanf("%d", &T);while (T--){cin >> n >> m >> k;for (int i = 0; i<k; i++)cin >> primes[i];for (int i = 0; i<k; i++)a[i] = Lucas(n, m, primes[i]);ll ans = China(k, a, primes);printf("%I64d\n", ans);}}

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