uva 11916 Emoogle Grid （BSGS）

UVA 11916

　　BSGS的一道简单题，不过中间卡了一下没有及时取模，其他这里的100000007是素数，所以不用加上拓展就能做了。

代码如下：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <map>

using namespace std;

template<class T> T gcd(T a, T b) { return b ? a : gcd(b, a % b);}
typedef long long LL;
void gcd(LL a, LL b, LL &d, LL &x, LL &y) {
    if (b) { gcd(b, a % b, d, y, x); y -= a / b * x;}
    else d = a, x = 1, y = 0;
}
const LL MOD = 100000007;
const int N = 555;

LL multi(LL a, LL p) {
    LL ret = 1;
    a %= MOD;
    while (p > 0) {
        if (p & 1) ret *= a, ret %= MOD;
        a *= a, a %= MOD, p >>= 1;
    }
    return ret;
}
map<LL, int> id;
vector<LL> rec[N];
LL mincol;

void bs(LL b, LL x, LL rt) {
    id.clear();
    for (int i = 0; i <= rt; i++) {
        if (id.find(b) != id.end()) break;
        id[b] = i;
        b *= x, b %= MOD;
    }
}

LL gs(LL b, LL x, LL r) {
    LL cur = b;
    //for (int i = 0; i < 50; i++) {
        //if (cur == r) return mincol + i + 1;
        //cur *= x, cur %= MOD;
    //}
    int rt = (int) ceil(sqrt((double) MOD)) + 1;
    bs(b, x, rt);
    LL st = multi(x, rt), p, q, d;
    cur = 1;
    for (int i = 0; i <= rt; i++) {
        gcd(cur, MOD, d, p, q);
        p %= MOD, p += MOD, p %= MOD, p *= r / d, p %= MOD;
        if (id.find(p) != id.end()) return (LL) i * rt + id[p] + mincol + 1;
        cur *= st, cur %= MOD;
    }
    return -1;
}

LL bf(LL b, LL x, LL r) {
    LL cur = b;
    for (int i = 0; i < MOD; i++) {
        if (cur == r) return mincol + i + 1;
        cur *= x, cur %= MOD;
    }
    return -1;
}
int main() {
    //freopen("in", "r", stdin);
    LL n, k, b, r, x, y, bres;
    int T, cas = 1;
    cin >> T;
    while (T-- && cin >> n >> k >> b >> r) {
        for (int i = 0; i < N; i++) rec[i].clear();
        id.clear();
        mincol = 1;
        for (int i = 0; i < b; i++) {
            cin >> x >> y;
            if (id.find(y) == id.end()) id[y] = id.size() - 1;
            rec[id[y]].push_back(x);
            mincol = max(mincol, x);
        }
        bres = 1;
        for (int i = 0, sz = id.size(); i < sz; i++) {
            rec[i].push_back(0);
            rec[i].push_back(mincol + 1);
            sort(rec[i].begin(), rec[i].end());
            for (int j = 1, szj = rec[i].size(); j < szj; j++) {
                if (rec[i][j] - rec[i][j - 1] - 2 >= 0) bres *= k * multi(k - 1, (LL) rec[i][j] - rec[i][j - 1] - 2) % MOD, bres %= MOD;
            }
        }
        LL tmp = k * multi(k - 1, mincol - 1);
        bres *= multi(tmp, n - id.size());
        bres %= MOD;
        cout << "Case " << cas++ << ": ";
        if (bres == r) { cout << mincol << endl; continue; }
        int cnt = 0;
        for (int i = 0, sz = id.size(); i < sz; i++) {
            rec[i].pop_back();
            if (rec[i][rec[i].size() - 1] == mincol) bres *= k, bres %= MOD, cnt++;
        }
        bres *= multi(k - 1, n - cnt);
        bres %= MOD;
        k = multi(k - 1, n);
        cout << gs(bres, k, r) << endl;
        //cout << bf(bres, k, r) << endl;
    }
    return 0;
}

——written by Lyon