UVA 12169 Disgruntled Judge (扩展欧几里得)

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原题地址:here

题意:有$3$个整数$x[1], a, b$ 满足递推式$x[i]=(a*x[i-1]+b)mod 10001$。由这个递推式计算出了长度为$2T$的数列，现在要求输入$x[1],x[3],......x[2T-1]$, 输出$x[2],x[4]......x[2T]. T<=100,0<=x<=10000.$如果有多种可能的输出，任意输出一个结果即可。

思路:因为题目中$a$的数据范围只有$1e4$,所以我们可以通过枚举$a$去确定$b$,那么如何去求$b$呢?
我们可以通过扩展欧几里得算法去求解线性同余方程.

#include <bits/stdc++.h>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,(rt<<1)+1
#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
const int mod = 10001;
ll x, y;
ll d1[maxn];
ll d2[maxn];
int n;
ll exgcd(ll a, ll b, ll &x, ll &y) { //求解ax+by=gcd(a,b)
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll ans = exgcd(b, a % b, x, y);
    ll temp = x;
    x = y;
    y = temp - (a / b) * y;
    return ans;
}


bool check(ll a, ll b) {
    for (int i = 1; i <= n; i++) {
        d2[i] = (d1[i] * a + b)%mod;
        if (i != n) {
            if (d1[i + 1] != (a * d2[i] + b) % mod) return 0;
        }
    }
    return 1;
}
int main() {

    while (~scanf("%d", &n)) {
        int f = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%lld", &d1[i]);
        }
        for (ll i = 0; i <= 10000; i++) {//枚举a
            ll gcd = exgcd((i + 1), 10001, x, y);

            ll sum = d1[2] - i * i * d1[1];//注意这里会爆int,因此要开long long
            if (sum % gcd) continue;
            ll b = x * sum / gcd;
            f = check(i, b);
            if (f) break;
        }
        for (int i = 1; i <= n; i++) printf("%lld\n", d2[i]);
    }
    return 0;
}