uva 11916 Emoogle Grid （BSGS）

最新推荐文章于 2021-06-11 14:59:53 发布

转载最新推荐文章于 2021-06-11 14:59:53 发布 · 129 阅读

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原文链接：http://www.cnblogs.com/LyonLys/p/uva_11916_Lyon.html

本文介绍了解决UVA11916问题的一种方法，使用BSGS（Baby-Step Giant-Step）算法进行求解。通过代码实现展示了如何有效地处理大数运算，并利用模意义下的快速幂计算，最终找到特定条件下序列的位置。

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：

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <vector>
  6 #include <cmath>
  7 #include <map>
  8 
  9 using namespace std;
 10 
 11 template<class T> T gcd(T a, T b) { return b ? a : gcd(b, a % b);}
 12 typedef long long LL;
 13 void gcd(LL a, LL b, LL &d, LL &x, LL &y) {
 14     if (b) { gcd(b, a % b, d, y, x); y -= a / b * x;}
 15     else d = a, x = 1, y = 0;
 16 }
 17 const LL MOD = 100000007;
 18 const int N = 555;
 19 
 20 LL multi(LL a, LL p) {
 21     LL ret = 1;
 22     a %= MOD;
 23     while (p > 0) {
 24         if (p & 1) ret *= a, ret %= MOD;
 25         a *= a, a %= MOD, p >>= 1;
 26     }
 27     return ret;
 28 }
 29 map<LL, int> id;
 30 vector<LL> rec[N];
 31 LL mincol;
 32 
 33 void bs(LL b, LL x, LL rt) {
 34     id.clear();
 35     for (int i = 0; i <= rt; i++) {
 36         if (id.find(b) != id.end()) break;
 37         id[b] = i;
 38         b *= x, b %= MOD;
 39     }
 40 }
 41 
 42 LL gs(LL b, LL x, LL r) {
 43     LL cur = b;
 44     //for (int i = 0; i < 50; i++) {
 45         //if (cur == r) return mincol + i + 1;
 46         //cur *= x, cur %= MOD;
 47     //}
 48     int rt = (int) ceil(sqrt((double) MOD)) + 1;
 49     bs(b, x, rt);
 50     LL st = multi(x, rt), p, q, d;
 51     cur = 1;
 52     for (int i = 0; i <= rt; i++) {
 53         gcd(cur, MOD, d, p, q);
 54         p %= MOD, p += MOD, p %= MOD, p *= r / d, p %= MOD;
 55         if (id.find(p) != id.end()) return (LL) i * rt + id[p] + mincol + 1;
 56         cur *= st, cur %= MOD;
 57     }
 58     return -1;
 59 }
 60 
 61 LL bf(LL b, LL x, LL r) {
 62     LL cur = b;
 63     for (int i = 0; i < MOD; i++) {
 64         if (cur == r) return mincol + i + 1;
 65         cur *= x, cur %= MOD;
 66     }
 67     return -1;
 68 }
 69 int main() {
 70     //freopen("in", "r", stdin);
 71     LL n, k, b, r, x, y, bres;
 72     int T, cas = 1;
 73     cin >> T;
 74     while (T-- && cin >> n >> k >> b >> r) {
 75         for (int i = 0; i < N; i++) rec[i].clear();
 76         id.clear();
 77         mincol = 1;
 78         for (int i = 0; i < b; i++) {
 79             cin >> x >> y;
 80             if (id.find(y) == id.end()) id[y] = id.size() - 1;
 81             rec[id[y]].push_back(x);
 82             mincol = max(mincol, x);
 83         }
 84         bres = 1;
 85         for (int i = 0, sz = id.size(); i < sz; i++) {
 86             rec[i].push_back(0);
 87             rec[i].push_back(mincol + 1);
 88             sort(rec[i].begin(), rec[i].end());
 89             for (int j = 1, szj = rec[i].size(); j < szj; j++) {
 90                 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;
 91             }
 92         }
 93         LL tmp = k * multi(k - 1, mincol - 1);
 94         bres *= multi(tmp, n - id.size());
 95         bres %= MOD;
 96         cout << "Case " << cas++ << ": ";
 97         if (bres == r) { cout << mincol << endl; continue; }
 98         int cnt = 0;
 99         for (int i = 0, sz = id.size(); i < sz; i++) {
100             rec[i].pop_back();
101             if (rec[i][rec[i].size() - 1] == mincol) bres *= k, bres %= MOD, cnt++;
102         }
103         bres *= multi(k - 1, n - cnt);
104         bres %= MOD;
105         k = multi(k - 1, n);
106         cout << gs(bres, k, r) << endl;
107         //cout << bf(bres, k, r) << endl;
108     }
109     return 0;
110 }

View Code

——written by Lyon

转载于:https://www.cnblogs.com/LyonLys/p/uva_11916_Lyon.html