SGU 455 Sequence analysis

http://acm.sgu.ru/problem.php?contest=0&problem=455

　　一道关于cycle detection技术的数论题。我用的是wiki中提到的第二种方法，跑出来的时间大约350ms。

　　题目说了数的范围是有符号的长整形，开始的时候我考虑到数据可能溢出，于是我就用了模乘法，可是交的时候在第6或第7个test的就TLE了。TLE了以后我就直接删去了模乘法，不过因为开始的时候限制计算的次数的判断的位置放的不对，所以在某些数据上会TLE或者wa。比较幸运的是，我随便测了一组数据就发现问题了。改过来以后就AC了～

原始代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxCnt = 2000000;

ll A, B, C;

ll multiMod(ll a, ll b, ll m) {
    ll ret = 0;

    while (b) {
        if (b & 1) {
            ret += a;
            ret %= m;
        }
        a <<= 1;
        a %= m;
        b >>= 1;
    }

    return ret;
}

ll cal(ll x) {
    return (A * x + x % B) % C;
//    if (x >= 0) return (multiMod(A, x, C) + x % B) % C;
//    else if (A >= 0) return (multiMod(x, A, C) + x % B) % C;
//    else return (multiMod(-x, -A, C) + x % B) % C;
}

ll brent(ll base, int &lam, int &mu) {
    ll ep, slow, fast;

    ep = lam = 1;
    slow = base;
    fast = cal(base);
    while (slow != fast) {
        if (ep == lam) { // get into a new power of two
            slow = fast;
            ep <<= 1;
            lam = 0;
        }
        fast = cal(fast);
        lam += 1;
        if (lam >= maxCnt) return -1;
    }
//    cout << slow << " " << fast << endl;
    // find the first repetition of lam
    mu = 0;
    slow = fast = base;
    for (int i = 0; i < lam; i++) {
        fast = cal(fast);
    }
//    cout << "lam " << lam << endl;
    while (fast != slow) {
        slow = cal(slow);
        fast = cal(fast);
        mu++;
    }

    if (mu + lam <= maxCnt) return mu + lam;
    else return -1;
}

int main() {
//    freopen("in", "r", stdin);

    while (cin >> A >> B >> C) {
        int first, cycle;

        cout << brent(1, cycle, first) << endl;
//        cout << first << ' ' << cycle << endl;
//        cout << A << ' ' << B << ' ' << C << endl;
//        for (ll i = 0, k = 1; i <= cycle + first; i++) {
//            cout << k << ' ';
//            k = cal(k);
//        }
//        cout << endl << endl;
    }

    return 0;
}

——written by Lyon