hrbeu 组合数 组合数对数取余

题目链接：http://acm.hrbeu.edu.cn/index.php?act=problem&id=1010&cid=25

其中P为非素数，所以有：

将C(n,m)%p=p0^C0```pi^Ci%p

#include<stdio.h>
#include<string.h>
#define LL long long
#define nmax 100001
int flag[nmax], prime[nmax];
int plen;
void mkprime() {
    int i, j;
    memset(flag, -1, sizeof(flag));
    for (i = 2, plen = 0; i < nmax; i++) {
        if (flag[i]) {
            prime[plen++] = i;
        }
        for (j = 0; (j < plen) && (i * prime[j] < nmax); j++) {
            flag[i * prime[j]] = 0;
            if (i % prime[j] == 0) {
                break;
            }
        }
    }
}
int getMin(int a, int b) {
    return a > b ? b : a;
}
int getNum(int n, int p) {
    int res;
    res = 0;
    while (n) {
        res += n / p;
        n /= p;
    }
    return res;
}
int modular_exp(int a, int b, int c) {
    LL res, temp;
    res = 1 % c, temp = a % c;
    while (b) {
        if (b & 1) {
            res = res * temp % c;
        }
        temp = temp * temp % c;
        b >>= 1;
    }
    return (int) res;
}
void solve(int a, int b, int c) {
    int i, temp;
    LL res;
    for (i = 0, res = 1; (i < plen) && (prime[i] <= a); i++) {
        temp = getNum(a, prime[i]) - getNum(b, prime[i])
                - getNum(a - b, prime[i]);
        res = res * modular_exp(prime[i], temp, c) % c;
    }
    printf("%lld\n", res);
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int t, a, b, c;
    mkprime();
    scanf("%d", &t);
    while (t--) {
        scanf("%d %d %d", &a, &b, &c);
        solve(a + b, getMin(a, b), c);
    }
    return 0;
}