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}.
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
1
9 5 2
3 5
Sample Output
6
题目要求一个大组合数模几个素数乘积的结果。
大组合那块能通过Lucas得到对每个素数模的结果。
然后再通过互质型的中国剩余定理可以得到最终结果。
不过我的模板中国剩余定理里面乘法部分会爆long long。
然后用大数优化了乘法部分，通过大数乘大数，模完后返回小数。
代码：
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <set> #include <queue> #include <vector> #define LL long long #define UNIT 10 using namespace std; const int maxK = 15; LL n, m, s[maxK], prime[maxK]; int k; struct Bignum { int val[105]; int len; Bignum() { memset(val, 0, sizeof(val)); len = 1; } Bignum operator=(const LL &a) { LL t, p = a; len = 0; while (p >= UNIT) { t = p - (p/UNIT)*UNIT; p = p / UNIT; val[len++] = t; } val[len++] = p; return *this; } Bignum operator*(const Bignum &a) const { Bignum x; int i, j, up; int x1, x2; for (i = 0; i < len; i++) { up = 0; for (j = 0; j < a.len; j++) { x1 = val[i]*a.val[j] + x.val[i+j] + up; if (x1 >= UNIT) { x2 = x1 - x1/UNIT*UNIT; up = x1 / UNIT; x.val[i+j] = x2; } else { up = 0; x.val[i+j] = x1; } } if (up != 0) x.val[i+j] = up; } x.len = i + j; while (x.val[x.len-1] == 0 && x.len > 1) x.len--; return x; } LL operator%(const LL &a) const { LL x = 0; for (int i = len-1; i >= 0; --i) x = ((x*UNIT)%a+val[i]) % a; return x; } }; LL mulMod(LL x, LL y, LL p) { LL ans = 0; Bignum xx, yy; xx = x; yy = y; xx = xx*yy; ans = xx%p; return ans; } LL quickMod(LL a, LL b, LL p) { LL ans = 1; a %= p; while (b) { if (b&1) { ans = ans*a%p; b--; } b >>= 1; a = a*a%p; } return ans; } LL C(LL n, LL m, LL p) { if (m > n) return 0; LL ans = 1; for(int i = 1; i <= m; i++) { LL a = (n+i-m)%p; LL b = i%p; ans = ans*(a*quickMod(b, p-2, p)%p)%p; } return ans; } LL Lucas(LL x, LL y, LL p) { if (y == 0) return 1; return C(x%p, y%p, p)*Lucas(x/p, y/p, p)%p; } //EXGCD //求解方程ax+by=d,即ax=d mod(b) //扩展可求逆元 //O(logn) void exgcd(LL a, LL b, LL &x, LL &y, LL &d) { if (b == 0) { x = 1; y = 0; d = a; } else { exgcd(b, a%b, y, x, d); y -= a/b*x; } } //中国剩余定理（互质） //其中a为除数数组,n为模数数组 LL CRT(LL *a, LL *n, int len) { LL N = 1, ans = 0; for (int i = 0; i < len; i++) { N *= n[i]; } for (int i = 0; i < len; i++) { LL m = N / n[i]; LL x, y, d; exgcd(m, n[i], x, y, d); x = (x%n[i] + n[i]) % n[i]; //ans = (ans + m*a[i]*x%N) % N; ans = (ans + mulMod(mulMod(m, a[i], N), x, N)) % N; } return ans; } void input() { scanf("%I64d%I64d%d", &n, &m, &k); for (int i = 0; i < k; ++i) scanf("%I64d", &prime[i]); } void work() { for (int i = 0; i < k; ++i) s[i] = Lucas(n, m, prime[i]); LL ans = CRT(s, prime, k); printf("%I64d ", ans); } int main() { //freopen("test.in", "r", stdin); int T; scanf("%d", &T); for (int times = 0; times < T; ++times) { input(); work(); } return 0; }