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

 

/*by SilverN*/
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
using namespace std;
const int mod=10001;
const int mxn=50000;
LL num[mxn];
LL x[mxn];
LL k,b;
int n;
LL exgcd(LL a,LL b,LL &x,LL &y){
    if(!b){
        x=1;y=0;
        return a;
    }
    LL tmp=exgcd(b,a%b,x,y);
    LL t=x;x=y;y=t-a/b*y;
    return tmp;
}
int main(){
    scanf("%d",&n);
    int i,j;
    for(i=1;i<=n;i++){
        scanf("%lld",&num[i]);
    }
    for(int a=0;a<mod;a++){
        LL tmp=num[2]-a*a*num[1];
        LL gcd=exgcd(a+1,mod,k,b);
        if(tmp%gcd)continue;
//        printf("test:%lld\n",tmp);
        b=k*tmp/gcd%mod;
        bool ok=1;
        x[1]=(num[1]*a+b)%mod;
        for(i=2;i<=n;i++){
            if(num[i]!=(x[i-1]*a+b)%mod){
                ok=0;
                break;
            }
            x[i]=(num[i]*a+b)%mod;
        }
        if(!ok)continue;
        //if ok
            for(i=1;i<=n;i++)printf("%lld\n",x[i]);
            break;
    }
    return 0;
}