HAZELNUT MUSEUM

摘要： Time Limit: 5 Seconds Memory Limit: 196608 KBYou are given a simple task. Given a sequence A[i] with N numbers. You have to perform Q operations on th 阅读全文

摘要： 康托展开与康托展开逆运算 ll fac[20]; //阶乘 void getFac() { fac[0] = 1; for(int i=1;i<20;i++) fac[i] = 1LL * fac[i-1] * i; } ll cantor(int *a,int len) //康托展开求a是全排列第 阅读全文