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。

 1 #include <cstdio>

 2 

 3 const int maxn = 10000 + 5;

 4 const int M = 10001;

 5 int T, x[maxn];

 6 

 7 int main()

 8 {

 9     //freopen("12169in.txt", "r", stdin);

10     

11     scanf("%d", &T);

12     for(int i = 1; i < 2*T; i += 2)

13         scanf("%d", &x[i]);

14     

15     bool ok;

16     for(int a = 0; a < M; ++a)

17     {

18         for(int b = 0; b < M; ++b)

19         {

20             ok = true;

21             for(int i = 2; i <= 2*T; i += 2)

22             {

23                 x[i] = (x[i-1]*a + b) % M;

24                 if(i < 2*T && x[i+1] != (x[i]*a + b) % M)

25                 {

26                     ok = false;

27                     break;

28                 }

29             }

30             if(ok) break;

31         }

32         if(ok) break;

33     }

34     

35     for(int i = 2; i <= 2*T; i += 2)

36         printf("%d\n", x[i]);

37     

38     return 0;

39 } 

代码君

 

第二种办法枚举a，根据x₁、x₃求b。

详见这里

 1 #include <cstdio>

 2 

 3 typedef long long LL;

 4 const int maxn = 10000 + 5;

 5 const LL M = 10001;

 6 int T;

 7 LL x[maxn];

 8 

 9 void gcd(LL a, LL b, LL& d, LL& x, LL& y)

10 {

11     if(!b) { d = a; x = 1; y = 0; }

12     else { gcd(b, a%b, d, y, x); y -= x*(a/b); }

13 }

14 

15 int main()

16 {

17     //freopen("12169in.txt", "r", stdin);

18     

19     scanf("%d", &T);

20     for(int i = 1; i < 2*T; i += 2)

21         scanf("%lld", &x[i]);

22     

23     bool ok;

24     for(LL a = 0; a < M; ++a)

25     {

26         LL t = x[3] - a*a*x[1];

27         LL g, k, b;

28         gcd(a+1, M, g, b, k);

29         if(t % g != 0) continue;

30         b *= t / g;

31         

32         ok = true;

33         for(int i = 2; i <= 2*T; i += 2)

34         {

35             x[i] = (x[i-1]*a+b) % M;

36             if(i < 2*T && x[i+1] != (x[i]*a+b) % M)

37             {

38                 ok = false;

39                 break;

40             }

41         }

42         if(ok) break;

43     }

44     

45     for(int i = 2; i <= 2*T; i += 2)

46         printf("%lld\n", x[i]);

47     

48     return 0;

49 } 

代码君