hdu 2769 uva 12169 Disgruntled Judge 拓展欧几里德

//数据是有多水 连 10^10的枚举都能过

关于拓展欧几里德：大概就是x1=y2,y1=x2-[a/b]y2,按这个规律递归到gcd(a,0)的形式，此时公因数为a，方程也变为a*x+0*y=gcd(a,0)的形式,显然解为x=1,y=0,然后再递归回去就能得到解（a*x+b*y=gcd(a,b)的解）

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<cmath>

 4 #include<algorithm>

 5 #include<cstring>

 6 #include<cstdlib>

 7 #include<queue>

 8 #include<vector>

 9 #include<map>

10 #include<stack>

11 #include<string>

12 

13 using namespace std;

14 

15 long long T;

16 long long x[201];

17 

18 void exGED(long long a,long long b,long long &d,long long &x,long long &y){

19     if (b==0){

20             x=1;

21             y=0;

22             d=a;

23     }

24     else{

25             exGED(b,a%b,d,x,y);

26             long long tmp=x;

27             x=y;

28             y=tmp-(a/b)*y;

29     }

30 }

31 

32 bool solve(long long a){

33     long long d,b,k;

34     long long tmp=x[3]-a*a*x[1];

35     exGED(a+1,10001,d,b,k);

36     if (tmp%d!=0) return false;

37     b=b*(tmp/d);

38     for (long long i=2;i<=2*T;i++){

39             if (i%2==0){

40                     x[i]=(x[i-1]*a+b)%10001;

41             }

42             else{

43                     if (x[i]!=((x[i-1]*a+b)%10001)){

44                             return false;

45                     }

46             }

47     }

48     for (long long i=2;i<=2*T;i+=2){

49             printf("%I64d\n",x[i]);

50     }

51     return true;

52 }

53 

54 int main(){

55     scanf("%I64d",&T);

56     for (long long i=1;i<2*T;i+=2){

57             scanf("%I64d",&x[i]);

58     }

59     //solve(1096);

60     for (long long a=0;a<=10000;a++){

61             if (solve(a)) break;

62     }

63     return 0;

64 }

65 /*

66 3

67 17

68 822

69 3014

70 */

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