标签:style blog http color os io 数据 for
可能由于后台数据的原因,这道题直接暴力枚举a,b进行判断也能过,不过跑的时间长,效率太差了。
14021006 | 12169 | Disgruntled Judge | Accepted | C++ | 0.876 | 2014-08-11 08:46:28 |
不说了,比较无脑。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<map> #include<set> #include<list> #include<string> #include<sstream> #include<ctime> using namespace std; #define _PI acos(-1.0) #define INF 1 << 10 #define esp 1e-6 typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> pill; /*=========================================== ===========================================*/ #define MAXD 200 + 10 int main(){ int T; int x[MAXD]; scanf("%d",&T); for(int i = 1 ; i < 2 * T ; i+= 2) scanf("%d",&x[i]); int ok,a,b; for(a = 0 ; a <= 10000 ; a++){ for(b = 0; b <= 10000 ; b++){ ok = 1; for(int i = 2 ; i <= 2 * T ; i++){ if(i & 1){ if(x[i] != ((a * x[i - 1] + b) % 10001)){ ok = 0; break; } } else{ x[i] = (a * x[i - 1] + b) % 10001; } } if(ok) break; } if(ok) break; } for(int i = 2 ; i <= 2 * T ; i+= 2) printf("%d\n",x[i]); return 0; }
如何通过a,x1,x3求出b呢。
x2 = (a * x1 + b) % 10001;
x3 = (a * x2 + b) % 10001;
联立2个式子
x3 = (a * (a * x1 + b) % 10001 + b ) % 10001;
x3 = (a * (a * x1 + b) + b) % 10001;
所以 x3 + 10001 * k = a * a * x1 + (a + 1) * b;
x3 - a * a * x1 = (a + 1) * b + 10001 * (-k);
这样就成了求 b 和 -k,满足这个式子,不就是扩展欧几里得的一般用法么?
14021484 | 12169 | Disgruntled Judge | Accepted | C++ | 0.012 | 2014-08-11 10:54:22 |
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<map> #include<set> #include<list> #include<string> #include<cmath> #include<sstream> #include<ctime> using namespace std; #define _PI acos(-1.0) #define INF 1 << 10 #define esp 1e-6 typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> pill; /*=========================================== ===========================================*/ #define MAXD 200 + 10 #define max_size 10001 void gcd(LL a , LL b ,LL &d, LL &x,LL &y){ if(!b){ d = a ; x = 1; y = 0; return ; } else{ gcd(b , a % b ,d , y , x); y -= x * (a / b); return ; } } int main(){ LL a,b,x[MAXD]; int T; scanf("%d",&T); for(int i = 1 ; i < 2 * T ; i += 2) scanf("%lld",&x[i]); for(a = 0; ; a++){ LL k , b , d; LL t = (x[3] - a * a * x[1]); gcd(max_size, a + 1, d , k, b); if(t % d) continue; b = b * t / d; int yes = 1; for(int i = 2 ; i <= 2 * T ; i ++){ if(i & 1){ if(x[i] != ((a * x[i - 1] + b) % max_size)){ yes = 0; break; } } else{ x[i] = (a * x[i - 1] + b) % max_size; } } if(yes) break; } for(int i = 2 ; i <= 2 * T ; i+=2) printf("%lld\n",x[i]); return 0; }
【UVA】12169-Disgruntled Judge(暴力or欧几里得),布布扣,bubuko.com
【UVA】12169-Disgruntled Judge(暴力or欧几里得)
标签:style blog http color os io 数据 for
原文地址:http://blog.csdn.net/u013451221/article/details/38497029