题意:
给出四个数T, a, b, x1,按公式生成序列 xi = (a*xi-1 + b) % 10001 (2 ≤ i ≤ 2T)
给出T和奇数项xi,输出偶数项xi
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。
1 #include <cstdio> 2 3 const int maxn = 10000 + 5; 4 const int M = 10001; 5 int T, x[maxn]; 6 7 int main() 8 { 9 //freopen("12169in.txt", "r", stdin); 10 11 scanf("%d", &T); 12 for(int i = 1; i < 2*T; i += 2) 13 scanf("%d", &x[i]); 14 15 bool ok; 16 for(int a = 0; a < M; ++a) 17 { 18 for(int b = 0; b < M; ++b) 19 { 20 ok = true; 21 for(int i = 2; i <= 2*T; i += 2) 22 { 23 x[i] = (x[i-1]*a + b) % M; 24 if(i < 2*T && x[i+1] != (x[i]*a + b) % M) 25 { 26 ok = false; 27 break; 28 } 29 } 30 if(ok) break; 31 } 32 if(ok) break; 33 } 34 35 for(int i = 2; i <= 2*T; i += 2) 36 printf("%d\n", x[i]); 37 38 return 0; 39 }
代码君
第二种办法枚举a,根据x1、x3求b。
详见这里
1 #include <cstdio> 2 3 typedef long long LL; 4 const int maxn = 10000 + 5; 5 const LL M = 10001; 6 int T; 7 LL x[maxn]; 8 9 void gcd(LL a, LL b, LL& d, LL& x, LL& y) 10 { 11 if(!b) { d = a; x = 1; y = 0; } 12 else { gcd(b, a%b, d, y, x); y -= x*(a/b); } 13 } 14 15 int main() 16 { 17 //freopen("12169in.txt", "r", stdin); 18 19 scanf("%d", &T); 20 for(int i = 1; i < 2*T; i += 2) 21 scanf("%lld", &x[i]); 22 23 bool ok; 24 for(LL a = 0; a < M; ++a) 25 { 26 LL t = x[3] - a*a*x[1]; 27 LL g, k, b; 28 gcd(a+1, M, g, b, k); 29 if(t % g != 0) continue; 30 b *= t / g; 31 32 ok = true; 33 for(int i = 2; i <= 2*T; i += 2) 34 { 35 x[i] = (x[i-1]*a+b) % M; 36 if(i < 2*T && x[i+1] != (x[i]*a+b) % M) 37 { 38 ok = false; 39 break; 40 } 41 } 42 if(ok) break; 43 } 44 45 for(int i = 2; i <= 2*T; i += 2) 46 printf("%lld\n", x[i]); 47 48 return 0; 49 }
代码君
时间: 2024-10-29 10:45:44