AC2012 - 博客园

2011年5月1日

摘要： // include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstream>#include <fstream>#include <iomanip>#include <bitset>#include <algorithm># 阅读全文

posted @ 2011-05-01 00:28 AC2012 阅读(159) 评论(0) 推荐(0)

2011年4月30日

1140

posted @ 2011-04-30 21:37 AC2012 阅读(213) 评论(0) 推荐(0)

1047

摘要： /*算是个模拟题吧*/// include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstream>#include <fstream>#include <iomanip>#include <bitset>#include <alg 阅读全文

posted @ 2011-04-30 21:03 AC2012 阅读(170) 评论(0) 推荐(0)

1032

摘要： /*将一个数分成若干段，成绩最大12 = 3 * 4 * 55 = 2 * 39 = 2 * 3 * 4210 = 5 * 6 * 710 = 2 * 3 * 511 = 2 * 4 * 55 = 2 * 36 = 2 * 47 = 3 * 48 = 3 * 59 = 2 * 3 * 410 = 2 * 3 * 511 = 2 * 4 * 512 = 3 * 4 * 513 = 3 * 4 * 614 = 2 * 3 * 4 * 5*/// include file#include <cstdio>#include <cstdlib>#include <cstri 阅读全文

posted @ 2011-04-30 20:19 AC2012 阅读(161) 评论(0) 推荐(0)

2142

摘要： /*扩展欧几里得a x1 x2b y1 y2(a*x1+b*y1) - (a*x2+b*y2) = da*x+b*y = D*/// include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstream>#include <fstream>#include & 阅读全文

posted @ 2011-04-30 17:21 AC2012 阅读(183) 评论(0) 推荐(0)

2891

摘要： /* m = a1*m1+r1 m = a2*m2+r2 m = a3*m3+r3 通过不断的合并方程求出最后的解，过程是这样的，以前两个方程为例 m === r1 mod a1 m === r2 mod a2 m = r1+a1*x r1+a1*x === r2 mod a2 r1+a1*x-r2 = a2*y a1*x-a2*y = r2-r1; r2 - r1 - a1*x = a2*y a1*x+a2*y = r2-r1 则可用扩展欧几里得算法求出x,y。则 m求出，不过这里只是对这两个方程而言,合并为新的方程，就有下式 m === (r1+a1*x) mod lcm(a1,a2)*/ 阅读全文

posted @ 2011-04-30 15:33 AC2012 阅读(361) 评论(0) 推荐(0)

1006

摘要： /*x-a = 23*a';x-b = 28*b';x-c = 33*c';x === a mod 23x === b mod 28x === c mod 33求x,这是中国剩余定理，这里应用要求m1,m2,m3两两互质x mod 23 = a mod 23x mod 28 = b mod 28x mod 33 = c mod 33需要用欧拉函数了phi(n)m = 23*28*33M1 = m/m1 = 28*33M2 = m/m2 = 23*33 M3 = m/m3 = 23*28M1' = M1^(phi(m1)-1) mod m1M2' = M2 阅读全文

posted @ 2011-04-30 14:00 AC2012 阅读(142) 评论(0) 推荐(0)

2011年4月29日

2115

摘要： /* mod = 2^k; A+C*k === B (mod mod) A+C*x - B = mod*y; C*x - mod*y = B-A; 求x a = C; b = -mod; c = B-A;*/// include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstre 阅读全文

posted @ 2011-04-29 23:31 AC2012 阅读(213) 评论(0) 推荐(0)

1061

摘要： /* (x+m*s)-(y+n*s) = k*L (k=0,1,2,...) (x-y) + (m-n)*s = k*L k*L + (n-m)*s = (x-y) 这是个方程，如果有解的话。可以使用扩展的欧几里得算法来求对扩展欧几里得写个小节： a*x+b*y = gcd(a,b) . 求解这个方程的解，一般用扩展欧几里得算法来解首先右边必须是gcd(a,b)的倍数才行，否则无解两边同时除以gcd(a,b); a'*x+b'*y = 1; 这样就可以用extGcd来解了,此时a'与b'互质 int extGcd(int a',int b' 阅读全文

posted @ 2011-04-29 23:02 AC2012 阅读(170) 评论(0) 推荐(0)

1183

摘要： /*这个题目的意思就是找出b的最大取值，然后枚举就可以了。默认b<=ce.g. */// include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstream>#include <fstream>#include <iomanip>#incl 阅读全文

posted @ 2011-04-29 17:37 AC2012 阅读(212) 评论(0) 推荐(0)

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