UVa 12169 (枚举+扩展欧几里得) Disgruntled Judge

题意：

给出四个数T, a, b, x₁,按公式生成序列 x_i = (a*x_i-1 + b) % 10001 (2 ≤ i ≤ 2T)

给出T和奇数项x_i，输出偶数项x_i

分析：

最简单的办法就是直接枚举a、b，看看与输入是否相符。

 #include <cstdio>

 const int maxn =  + ;
 const int M = ;
 int T, x[maxn];

 int main()
 {
     //freopen("12169in.txt", "r", stdin);

     scanf("%d", &T);
     for(int i = ; i < *T; i += )
         scanf("%d", &x[i]);

     bool ok;
     for(int a = ; a < M; ++a)
     {
         for(int b = ; b < M; ++b)
         {
             ok = true;
             for(int i = ; i <= *T; i += )
             {
                 x[i] = (x[i-]*a + b) % M;
                 if(i < *T && x[i+] != (x[i]*a + b) % M)
                 {
                     ok = false;
                     break;
                 }
             }
             if(ok) break;
         }
         if(ok) break;
     }

     for(int i = ; i <= *T; i += )
         printf("%d\n", x[i]);

     return ;
 } 

代码君

第二种办法枚举a，根据x₁、x₃求b。

详见这里

 #include <cstdio>

 typedef long long LL;
 const int maxn =  + ;
 const LL M = ;
 int T;
 LL x[maxn];

 void gcd(LL a, LL b, LL& d, LL& x, LL& y)
 {
     if(!b) { d = a; x = ; y = ; }
     else { gcd(b, a%b, d, y, x); y -= x*(a/b); }
 }

 int main()
 {
     //freopen("12169in.txt", "r", stdin);

     scanf("%d", &T);
     for(int i = ; i < *T; i += )
         scanf("%lld", &x[i]);

     bool ok;
     for(LL a = ; a < M; ++a)
     {
         LL t = x[] - a*a*x[];
         LL g, k, b;
         gcd(a+, M, g, b, k);
         if(t % g != ) continue;
         b *= t / g;

         ok = true;
         for(int i = ; i <= *T; i += )
         {
             x[i] = (x[i-]*a+b) % M;
             if(i < *T && x[i+] != (x[i]*a+b) % M)
             {
                 ok = false;
                 break;
             }
         }
         if(ok) break;
     }

     for(int i = ; i <= *T; i += )
         printf("%lld\n", x[i]);

     return ;
 } 

代码君