快速读入 线性求逆元

#多校赛2017#
#模板#

出题人给的黑科技：

namespace fastIO {
    #define BUF_SIZE 100000
    //fread -> read
    bool IOerror = 0;
    inline char nc() {
        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
        if(p1 == pend) {
            p1 = buf;
            pend = buf + fread(buf, 1, BUF_SIZE, stdin);
            if(pend == p1) {
                IOerror = 1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
    }
    inline void read(int &x) {
        char ch;
        while(blank(ch = nc()));
        if(IOerror)
            return;
        for(x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
    }
    #undef BUF_SIZE
};

线性求逆元：

    iact[1] = 1;
    for(int i = 2; i < maxn; ++i)
        iact[i] = mod - (int)(mod / i * (LL)iact[mod % i] % mod);
    fact[0] = iact[0] = 1;
    for(int i = 1; i < maxn; ++i)
    {
        fact[i] = (LL)fact[i - 1] * i % mod;
        iact[i] = (LL)iact[i - 1] * iact[i] % mod; //此处为求阶乘的逆元
    }