UVa12169 Disgruntled Judge

x2 = (a * x1 + b) % 10001;
x3 = (a * x2 + b) % 10001;

∴x3 = (a * (a * x1 + b) % 10001 + b ) % 10001;

∴(a + 1) * b + 10001 * (-k) = x3 - a * a * x1

由于a的范围是1~10000，所以可以枚举a，解出b，暴力判断这组a和b能否适用于所有的x

 1 /*by SilverN*/
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #define LL long long
 8 using namespace std;
 9 const int mod=10001;
10 const int mxn=50000;
11 LL num[mxn];
12 LL x[mxn];
13 LL k,b;
14 int n;
15 LL exgcd(LL a,LL b,LL &x,LL &y){
16     if(!b){
17         x=1;y=0;
18         return a;
19     }
20     LL tmp=exgcd(b,a%b,x,y);
21     LL t=x;x=y;y=t-a/b*y;
22     return tmp;
23 }
24 int main(){
25     scanf("%d",&n);
26     int i,j;
27     for(i=1;i<=n;i++){
28         scanf("%lld",&num[i]);
29     }
30     for(int a=0;a<mod;a++){
31         LL tmp=num[2]-a*a*num[1];
32         LL gcd=exgcd(a+1,mod,k,b);
33         if(tmp%gcd)continue;
34 //        printf("test:%lld\n",tmp);
35         b=k*tmp/gcd%mod;
36         bool ok=1;
37         x[1]=(num[1]*a+b)%mod;
38         for(i=2;i<=n;i++){
39             if(num[i]!=(x[i-1]*a+b)%mod){
40                 ok=0;
41                 break;
42             }
43             x[i]=(num[i]*a+b)%mod;
44         }
45         if(!ok)continue;
46         //if ok
47             for(i=1;i<=n;i++)printf("%lld\n",x[i]);
48             break;
49     }
50     return 0;
51 }