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题意:找出m个不同的n位2进制数,异或值中前v个为1,其余为0的方案数,答案 % 10567201.. 思路:比赛时第一感觉是用容斥原理做的,然后推呀推,搞了2个小时还是错了..赛后才知道递推才是正解(也许容斥是可以的,是我太弱了,推不出吧) 因为异或的特性,所以这m个数异或为x(前v个为1,其余为0的m位数),相当于这m个数异或x为0,.. 也就是说如果知道m-1个数,第m个数也唯一被确定了.. 假设f[m]为m个数的方案数,那么不考虑重复的情况下,f[m] = C(2^n, m-1) 那么如…

HDOJ 4770 Lights Against Dudely

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poj1222 EXTENDED LIGHTS OUT 高斯消元||枚举

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HDU 4770 Lights Against DudelyLights

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Traffic Lights

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