bigsnowstorm的专栏

//升序排列；先确定最大的数 void bubble_sort(int a[], int size) { bool swapped = false; for(int i = 0; i < size-1 ; i++) //外层进行size-1次循环，如有10个数，外层进行进行9次循环 { swapped = false; for(int j=0; j < size-1-i ; j++)

#include int cal_max(const int &a, const int &b) { int c; c = (a+b+abs(a-b))/2; return c; } int cal_max_2(const int &a, const int &b) { bool c=false; c = (bool)((unsigned int)(a-b) >> (sizeof(

书中首先介绍了两种简单的方法： One way to solve this problem would be to read the n numbers into an array, sort the array in decreasing order by some simple algorithm such as bubble sort, and then return the elemen

2.1. Mathematical Background Throughout the book we will use the following four definitions: DEFINITION: T(n) = O(f(n)) if there are constants c and n0 such that T(n)cf (n) when n>= n0. DEFINITI

一、递归的四条基本法则 When writing recursive routines, it is crucial to keep in mind the four basic rules of recursion: 1. Base cases.You must always have some base cases, which can be solved without recursio

//用递归实现n是2的几次方 int log2(int n) { if(!(n&(n-1))) { if(1 == n) { return 0; } else { return 1+log2(n>>1); } } else { return 0; } }

int find_sec_max(int *a, int size) { int max, sec_max; if(a[0] > a[1]) { max = a[0]; sec_max = a[1]; } else { max = a[1]; sec_max = a[0]; } for(int i=2; i < size; i++) { if(a[i] >

//给定一个十进制整数N,求出从1到N的所有整数中出现”1”的个数。 int cal_num_of_1(int n) { int count = 0; int temp; for(int i = 1; i<n+1; i++) { temp = i; while(temp>0) { if(temp%10 == 1) { count++; } tem