UVa12169 - Disgruntled Judge
/*UVa12169 - Disgruntled Judge --a,b都在10000以内,这样可以枚举a的值,然后根据x1和x3计算出b的值,再遍历数组X判断他们是否合法 --很容易得到x3=(a*a*x1+(a+1)*b)%10001;变形一下可以得到:(a+1)*b+10001*k=x3-a*a*x1;直接利用扩展欧几里得算法 求解b在[0,10000]的唯一解。 */ #define _CRT_SECURE_NO_DEPRECATE #include<iostream> using namespace std; const int maxn =100 + 10; const int mod = 10001; typedef long long LL; LL F[maxn * 2]; //扩展欧几里德算法,求解ax+by=gcd(a,b)系数 void extendEuclid(LL a, LL b, LL &d, LL &x, LL&y){ if (!b){ d = a; x = 1; y = 0; } else{ extendEuclid(b, a%b, d, y, x); y -= x*(a / b); } } //求ax+by=n的一个解(x0,y0) //所有解为(x0+t*b,y0-t*a),t是任意整数 LL solveLiner(LL a, LL b, LL n, LL&x, LL&y){ LL d; extendEuclid(a, b, d, x, y); if (n%d)return 0; //无解 x = (n / d)*x; y = (n / d)*y; return 1; } inline LL answer(LL a, LL b, LL x){ return (a*x + b) % mod; } int main(){ int T,i; LL a, b,k; while (~scanf("%d", &T)){ for (i = 1; i <= 2 * T - 1; i += 2) scanf("%d", &F[i]); for (a = 0; a < mod; a++){ solveLiner(a + 1, mod, F[3] - a*a*F[1], b, k); b = (b%mod + mod) % mod; //计算出b在[0,10000]的解 for (i = 2; i <= 2 * T; i++){ LL temp = answer(a, b, F[i - 1]); if (i & 1 && temp != F[i])break; F[i] = temp; } if (i == 2 * T + 1)break; } for (i = 2; i <= 2 * T; i += 2) printf("%I64d\n", F[i]); } return 0; }