uva12169-disgruntled Judge (modulo operation)

If you know a, you can calculate the X2, according to X3= (ax2+b) mod10001 calculate B.
The entire sequence can then be computed in O (T) time.
If a contradiction is found and entered in the calculation, a is illegal because a is an integer of 0~10000, even if all a is enumerated, the time efficiency is high enough.
Enumeration A, using X1,X3 to find B, the relationship between all x can not satisfy a, a, a.

How to find B through a,x1,x3.
x2 = (A * x1 + b)% 10001;
x3 = (A * x2 + b)% 10001;
The 2 formulas of the United State
x3 = (A * (A * x1 + b)% 10001 + b)% 10001;
x3 = (A * (A * x1 + b) + b)% 10001;
So x3 + 10001 * k = A * A * x1 + (A + 1) * b;
X3-a * A * x1 = (A + 1) * b + 10001 * (-K);
So it's a matter of asking for B and K, to satisfy this equation, not to extend the general usage of Euclid?

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include < vector> #include <stack> #include <queue> #include <map> #include <set> #include <list> #include <string> #include <cmath> #include <sstream> #include <ctime>using namespace std;# Define _PI ACOs ( -1.0) #define INF 1 << 10#define esp 1e-6typedef long long ll;typedef unsigned long long ull;typedef    pair<int,int> pill; #define MAXD + 10#define max_size 10001void gcd (LL a,ll b,ll &d,ll &x,ll &y) {        if (!b) {d=a;        x=1;y=0;    Return        } else {gcd (b, a%b, D, y, x);        Y-= x * (A/b);    return;    }}int Main () {LL a,b,x[maxd];    int T;    scanf ("%d", &t);    for (int i=1;i<2*t;i+=2) scanf ("%lld", &x[i]); for (a=0;;        a++) {LL k, B, D;        LL t = (x[3]-a*a*x[1]);        GCD (Max_size, a+1, D, K, b);        if (t% d) continue;        b = b * T/D; int Yes = 1; for (int i=2;i<=2*t;i++) {if (i&1) {if (x[i]!= ((a*x[i-1]+b)%max_size)) {y                    es=0;                Break            }} else {x[i]= (a*x[i-1]+b)%max_size;    }} if (yes) break;    } for (int i=2;i<=2*t;i+=2) printf ("%lld\n", X[i]); return 0;}

uva12169-disgruntled Judge (modulo operation)