MOD - Power Modulo Inverted(SPOJ3105) + Clever Y(POJ3243) + Hard Equation (Gym 101853G ) + EXBSGS

思路：

　　前两题题面相同，代码也相同，就只贴一题的题面了。这三题的意思都是求A^X==B(mod P），P可以不是素数，EXBSGS板子题。

SPOJ3105题目链接：https://www.spoj.com/problems/MOD/

POJ3243题目链接:http://poj.org/problem?id=3243

题目:

代码实现如下：

 1 #include <set>
 2 #include <map>
 3 #include <queue>
 4 #include <stack>
 5 #include <cmath>
 6 #include <bitset>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstdlib>
11 #include <cstring>
12 #include <iostream>
13 #include <algorithm>
14 #include <unordered_map>
15 using namespace std;
16
17 typedef long long LL;
18 typedef pair<LL, LL> pLL;
19 typedef pair<LL, int> pli;
20 typedef pair<int, LL> pil;;
21 typedef pair<int, int> pii;
22 typedef unsigned long long uLL;
23
24 #define lson i<<1
25 #define rson i<<1|1
26 #define lowbit(x) x&(-x)
27 #define bug printf("*********\n");
28 #define debug(x) cout<<"["<<x<<"]" <<endl;
29 #define FIN freopen("D://code//in.txt", "r", stdin);
30 #define IO ios::sync_with_stdio(false),cin.tie(0);
31
32 const double eps = 1e-8;
33 const int mod = 1e9 + 7;
34 const int maxn = 1e6 + 7;
35 const double pi = acos(-1);
36 const int inf = 0x3f3f3f3f;
37 const LL INF = 0x3f3f3f3f3f3f3f3f;
38
39 int x, z, k;
40 unordered_map<LL, int> mp;
41
42 int Mod_Pow(int x, int n, int mod) {
43     int res = 1;
44     while(n) {
45         if(n & 1) res = (LL)res * x % mod;
46         x = (LL)x * x % mod;
47         n >>= 1;
48     }
49     return res;
50 }
51
52 int gcd(int  a, int b) {
53     return b == 0 ? a : gcd(b, a % b);
54 }
55
56 int EXBSGS(int A, int B, int C) {
57     A %= C, B %= C;
58     if(B == 1) return 0;
59     int cnt = 0;
60     LL t = 1;
61     for(int g = gcd(A, C); g != 1; g = gcd(A, C)) {
62         if(B % g) return -1;
63         C /= g, B /= g, t = t * A / g % C;
64         cnt++;
65         if(B == t) return cnt;
66     }
67     mp.clear();
68     int m = ceil(sqrt(1.0*C));
69     LL base = B;
70     for(int i = 0; i < m; i++) {
71         mp[base] = i;
72         base = base * A % C;
73     }
74     base = Mod_Pow(A, m, C);
75     LL nw = t;
76     for(int i = 1; i <= m; i++) {
77         nw = nw * base % C;
78         if(mp.count(nw)) {
79             return i * m - mp[nw] + cnt;
80         }
81     }
82     return -1;
83 }
84
85 int main() {
86     //FIN;
87     while(~scanf("%d%d%d", &x, &z, &k)) {
88         if(x == 0 && z == 0 && k == 0) break;
89         int ans = EXBSGS(x, k, z);
90         if(ans == -1) printf("No Solution\n");
91         else printf("%d\n", ans);
92     }
93     return 0;
94 }

Gym 101853G题目链接：http://codeforces.com/gym/101853/problem/G

代码实现如下：

 1 #include <set>
 2 #include <map>
 3 #include <queue>
 4 #include <stack>
 5 #include <cmath>
 6 #include <bitset>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstdlib>
11 #include <cstring>
12 #include <iostream>
13 #include <algorithm>
14 #include <unordered_map>
15 using namespace std;
16
17 typedef long long LL;
18 typedef pair<LL, LL> pLL;
19 typedef pair<LL, int> pli;
20 typedef pair<int, LL> pil;;
21 typedef pair<int, int> pii;
22 typedef unsigned long long uLL;
23
24 #define lson i<<1
25 #define rson i<<1|1
26 #define lowbit(x) x&(-x)
27 #define bug printf("*********\n");
28 #define debug(x) cout<<"["<<x<<"]" <<endl;
29 #define FIN freopen("D://code//in.txt", "r", stdin);
30 #define IO ios::sync_with_stdio(false),cin.tie(0);
31
32 const double eps = 1e-8;
33 const int mod = 1e9 + 7;
34 const int maxn = 1e6 + 7;
35 const double pi = acos(-1);
36 const int inf = 0x3f3f3f3f;
37 const LL INF = 0x3f3f3f3f3f3f3f3f;
38
39 int t, a, b, m;
40 unordered_map<LL, int> mp;
41
42 LL Mod_Pow(LL x, LL n, LL mod) {
43     LL res = 1;
44     while(n) {
45         if(n & 1) res = res * x % mod;
46         x = x * x % mod;
47         n >>= 1;
48     }
49     return res;
50 }
51
52 int gcd(int a, int b) {
53     return b == 0 ? a : gcd(b, a % b);
54 }
55
56 LL EXBSGS(int A, int B, int C) {
57     A %= C, B %= C;
58     if(B == 1) return 0;
59     int cnt = 0;
60     LL t = 1;
61     for(int g = gcd(A, C); g != 1; g = gcd(A, C)) {
62         if(B % g) return -1;
63         C /= g, B /= g;
64         t = t * A / g % C;
65         cnt++;
66         if(B == t) return cnt;
67     }
68     mp.clear();
69     int m = ceil(sqrt(1.0 * C));
70     LL base = B;
71     for(int i = 0; i < m; i++) {
72        mp[base] = i;
73        base = base * A % C;
74     }
75     base = Mod_Pow(A, m, C);
76     LL nw = t;
77     for(int i = 1; i <= m + 1; i++) {
78         nw = base * nw % C;
79         if(mp.count(nw)) {
80             return i * m - mp[nw] + cnt;
81         }
82     }
83     return -1;
84 }
85
86 int main() {
87     scanf("%d", &t);
88     while(t--) {
89         scanf("%d%d%d", &a, &b, &m);
90         LL ans = EXBSGS(a, b, m);
91         printf("%lld\n", ans);
92     }
93     return 0;
94 }

转载于:https://www.cnblogs.com/Dillonh/p/9512030.html

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