UVa 12169 (枚举+扩展欧几里得) Disgruntled Judge

题意：

给出四个数T, a, b, x₁,按公式生成序列 x_i = (a*x_i-1 + b) % 10001 (2 ≤ i ≤ 2T)

给出T和奇数项x_i，输出偶数项x_i

分析：

最简单的办法就是直接枚举a、b，看看与输入是否相符。

#include <cstdio>

const int maxn = 10000 + 5;
const int M = 10001;
int T, x[maxn];

int main()
{
    //freopen("12169in.txt", "r", stdin);
    
    scanf("%d", &T);
    for(int i = 1; i < 2*T; i += 2)
        scanf("%d", &x[i]);
    
    bool ok;
    for(int a = 0; a < M; ++a)
    {
        for(int b = 0; b < M; ++b)
        {
            ok = true;
            for(int i = 2; i <= 2*T; i += 2)
            {
                x[i] = (x[i-1]*a + b) % M;
                if(i < 2*T && x[i+1] != (x[i]*a + b) % M)
                {
                    ok = false;
                    break;
                }
            }
            if(ok) break;
        }
        if(ok) break;
    }
    
    for(int i = 2; i <= 2*T; i += 2)
        printf("%d
", x[i]);
    
    return 0;
} 

第二种办法枚举a，根据x₁、x₃求b。

详见这里

#include <cstdio>

typedef long long LL;
const int maxn = 10000 + 5;
const LL M = 10001;
int T;
LL x[maxn];

void gcd(LL a, LL b, LL& d, LL& x, LL& y)
{
    if(!b) { d = a; x = 1; y = 0; }
    else { gcd(b, a%b, d, y, x); y -= x*(a/b); }
}

int main()
{
    //freopen("12169in.txt", "r", stdin);
    
    scanf("%d", &T);
    for(int i = 1; i < 2*T; i += 2)
        scanf("%lld", &x[i]);
    
    bool ok;
    for(LL a = 0; a < M; ++a)
    {
        LL t = x[3] - a*a*x[1];
        LL g, k, b;
        gcd(a+1, M, g, b, k);
        if(t % g != 0) continue;
        b *= t / g;
        
        ok = true;
        for(int i = 2; i <= 2*T; i += 2)
        {
            x[i] = (x[i-1]*a+b) % M;
            if(i < 2*T && x[i+1] != (x[i]*a+b) % M)
            {
                ok = false;
                break;
            }
        }
        if(ok) break;
    }
    
    for(int i = 2; i <= 2*T; i += 2)
        printf("%lld
", x[i]);
    
    return 0;
}