2015上海网络赛 HDU 5478 Can you find it 数学

HDU 5478 Can you find it

题意略。

思路：先求出n = 1 时候满足条件的(a,b), 最多只有20W对，然后对每一对进行循环节判断即可

 #include <iostream>
 #include <cstdio>
 #include <fstream>
 #include <algorithm>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <string>
 #include <cstring>
 #include <map>
 #include <stack>
 #include <set>
 #define LL long long
 #define eps 1e-8
 #define INF 0x3f3f3f3f
 #define MAXN 200005
 using namespace std;
 LL C, k1, b1, k2;
 LL quick_mod(LL x, LL y, LL mod){
     if (y == ) return ;
     LL res = quick_mod(x, y >> , mod);
     res = res * res % mod;
     if (y & ){
         res = res * x % mod;
     }
     return res;
 }
 LL a[MAXN][];
 LL b[MAXN];
 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     //freopen("out.txt", "w", stdout);
 #endif // OPEN_FILE
     int cas = ;
     while (~scanf("%I64d%I64d%I64d%I64d", &C,&k1, &b1, &k2)){
         printf("Case #%d:\n", cas++);
         int t = ;
         for (int i = ; i < C; i++){
             LL m = quick_mod((LL)i, k1 + b1, C);
             if (m == ) continue;
             a[++t][] = i;
             a[t][] = C - m;
             b[t] = m;
         }
         bool noans = true;
         for (LL i = ; i <= t; i++){
             LL pa = b[i];
             LL pa_k1 = quick_mod(a[i][], k1, C);
             LL pb_k2 = quick_mod(a[i][], k2, C);
             LL pb = a[i][];
             LL xx = pa;
             LL yy = pb;
         //    pb_k2 = quick_mod(pb_k2, 10000, C);
             LL x = pa_k1, y = pb_k2;
             bool flag = false;
             for (int j = ; j <= ; j++){
                 pa = pa * x % C;
                 pb = pb * y % C;
                 if (pa == xx && pb == yy){
                     break;
                 }
                 if ((pa + pb) % C != ){
                     flag = true;
                     break;
                 }
             }
             if (flag) continue;
             printf("%I64d %I64d\n", a[i][], a[i][]);
             noans = false;
         }
         if (noans){
             printf("-1\n");
         }
     }
 }