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HDU 5446 Unknown Treasure
Accept: 0 Submit: 0
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Problem Description
On the way to the next secret treasure hiding place, the mathematician discovered a cave unknown to the map. The mathematician entered the cave because it is there. Somewhere deep in the cave, she found a treasure chest with a combination lock and some numbers on it. After quite a research, the mathematician found out that the correct combination to the lock would be obtained by calculating how many ways are there to pick m different apples among n of them and modulo it with M. M is the product of several different primes.
Input
On the first line there is an integer T(T≤20) representing the number of test cases.
Each test case starts with three integers n,m,k(1≤m≤n≤10^18,1≤k≤10) on a line where k is the number of primes. Following on the next line are k different primes p1,...,pk. It is guaranteed that M=p1?p2???pk≤10^18 and pi≤10^5 for every i∈{1,...,k}.
Output
For each test case output the correct combination on a line.
Sample Input
9 5 2
3 5
Sample Output
Problem Idea
解题思路:
【题意】
求解
【类型】
lucas定理+中国剩余定理
【分析】
由于M为k个不同质数的乘积
所以M不一定为质数,这就导致我们无法直接使用lucas定理来求解组合数取模
那我们该怎么办呢?
先来回顾一下中国剩余定理
我们知道,对于一元模线性方程组
它的解可以写成x=a+kb(k≥0)的形式,并且
再回过头来看这道题,
在lucas定理不能直接使用的情况下,我们是否能够缩小
答案显然是可以的
我们可以通过中国剩余定理,将
所以题目到此就转化为了求一元模线性方程组的最小解,即a
那么等价的一元模线性方程组是多少呢?如下:
于是乎,此题到此结束
【时间复杂度&&优化】
题目链接→HDU 5446 Unknown Treasure
Source Code
/*Sherlock and Watson and Adler*/
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<bitset>
#include<cmath>
#include<complex>
#include<string>
#include<algorithm>
#include<iostream>
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define bitnum(a) __builtin_popcount(a)
using namespace std;
const int N = 15;
const int M = 100005;
const int inf = 1000000007;
const int mod = 7;
__int64 quick_mod(__int64 a,__int64 b,__int64 p)
{__int64 ans=1;a%=p;while(b){if(b&1){ans=ans*a%p;b--;}b>>=1;a=a*a%p;}return ans;
}
__int64 C(__int64 n,__int64 m,__int64 p)
{if(m>n) return 0;__int64 ans=1;for(int i=1;i<=m;i++){__int64 a=(n+i-m)%p;__int64 b=i%p;ans=ans*(a*quick_mod(b,p-2,p)%p)%p;}return ans;
}
__int64 Lucas(__int64 n,__int64 m,__int64 p)
{if(m==0) return 1;return C(n%p,m%p,p)*Lucas(n/p,m/p,p)%p;
}
int group;
LL n[N],a[N];
LL Egcd(LL a,LL b,LL &x,LL &y)
{if(b==0){x=1,y=0;return a;}LL d,tp;d=Egcd(b,a%b,x,y);tp=x;x=y;y=tp-a/b*y;return d;
}
LL solve()
{int i;bool flag = false;LL n1 = n[0], n2, b1 = a[0], b2, bb, d, t, k, x, y;for (i = 1; i < group; i++){n2 = n[i], b2 = a[i];bb = b2 - b1;d = Egcd (n1, n2, x, y);if (bb % d) //模线性解k1时发现无解{flag = true;break;}k = bb / d * x; //相当于求上面所说的k1【模线性方程】t = n2 / d;if (t < 0) t = -t;k = (k % t + t) % t; //相当于求上面的K`b1 = b1 + n1*k;n1 = n1 / d * n2;}if(flag)return -1; //无解
/******************求正整数解******************/if(b1==0) //如果解为0，而题目要正整数解，显然不行b1=n1; //n1刚好为所有ni的最小公倍数，就是解了
/******************求正整数解******************/return b1; //形成的解：b1, b1+n1, b1+2n1,..., b1+xni...
}
int main()
{int i,t;__int64 p,q,M;scanf("%d",&t);while(t--){M=1;scanf("%I64d%I64d%d",&p,&q,&group);for(i=0;i<group;i++){scanf("%I64d",&n[i]);M*=n[i];a[i]=Lucas(p,q,n[i]);}printf ("%I64d\n",solve()%M);}return 0;
}