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Django-Database 之 Many-To-Many关系

这里对Many-To-Many即多对多的映射关系以详细事例来分析Django中Database操作多对多映射关系的一些基本用法和注意事项

2018-08-02

6450

Coderbyte-Challenger之Letter Capitalize（单词字母大写）

Have the function LetterCapitalize(str) take the str parameter being passed and capitalize the first letter of each word. Words will be separated by only one space.

2018-08-02

7450

PAT Advanced 1005

Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.

2018-08-02

2820

PAT Advanced 1066

An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illustrate the rotation rules.

2018-08-02

2330

PAT Advanced 1067

1067. Sort with Swap(0,*) (25) 时间限制 100 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Given any permutation of the numbers {0, 1, 2,..., N-1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY oper

2018-08-02

2660

PAT Advanced 1042

Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techniques are seen as weak, and in order to avoid "inside jobs" where employees collaborate with gamblers by performing inadequate shuffles, many casinos employ automatic shuffling machines. Your task is to simulate a shuffling machine.

2018-08-02

2600

PAT Advanced 1043

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

2018-08-02

2580

PAT Advanced 1001

Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

2018-08-02

3120

#include <stdio.h> int main() { int n, e, flag = 0; while (scanf("%d%d", &n, &e) != EOF) { if( n*e ) { if(flag) printf(" "); else flag = 1;

2018-08-02

5180

#include<stdio.h> #include<string.h> /*此题和1008类似*/ void reverse(char *a,int left,int right) { while(left<right) { char temp=a[left]; a[left]=a[right]; a[right]=temp; left++;right--; } } int main() { char a[81]; gets(a); reverse(a,0,strlen(a)

2018-08-02

3390

#include<stdio.h> #include<stdlib.h> /*三次逆置*/ void reverse(int *a,int left,int right) { while(left<right) { int temp=a[left]; a[left]=a[right]; a[right]=temp; left++;right--; } } int main() { int n,k,i; scanf("%d %d",&n,&k); int *a=(int *)m

2018-08-02

4640

#include<stdio.h> #include<math.h> int a[100000]={0}; int main() { int n,i,j,flag,count=0; scanf("%d",&n); a[2]=a[3]=1; for(i=5;i<=n;i=i+2) { flag=1; for(j=3;j<=sqrt(i);j++) { if(a[j]==1) if(i%j==0) flag=0; } a[i]=flag; } i=3;

2018-08-02

5340

#include<stdio.h> /*简单的模拟题*/ int main() { int n,a[3]={0},count=2,i; scanf("%d",&n); while(n/10 != 0) { a[count--]=n%10; n=n/10; } a[count]=n%10; if(a[0]!=0) for(i=0;i<a[0];i++) printf("%c",'B'); if(a[1]!=0) for(i=0;i<a[1];i++) printf(

2018-08-02

5560

#include<stdio.h> #include<stdlib.h> int b[101]={0}; int main() { int n,i,temp,count=0,j,temp2; scanf("%d",&n); int *a=(int *)malloc(n*sizeof(int)); for(i=0;i<n;i++) scanf("%d",&a[i]); /*sort*/ for(i=0;i<n;i++) { int min=i; for(j=i;j<n;j++) {

2018-08-02

9600

#include<stdio.h> char a[10][5]={"ling","yi","er","san","si","wu","liu","qi","ba","jiu"}; /*递归的从左至右输出各个位*/ void b(int sum,int count) { count++; if(sum/10 !=0 ) b(sum/10,count); if(count==1) printf("%s",a[sum%10]); else printf("%s ",a[sum%10]); }

2018-08-02

6690

/*简单的模拟题*/ #include<stdio.h> int main() { int n,count=0; scanf("%d",&n); while(n!=1) { n=((n&1

2018-08-02

6030

PAT Advanced 1065

Given three integers A, B and C in [-263, 263], you are supposed to tell whether A+B > C.

2018-08-02

2540

较快速在一个数组中查找最大值和最小值(2)

/*分治法*/ #include<iostream> #include<string> #include<vector> #include<fstream> using namespace std; int a[16]={1,3,5,7,9,11,14,2,4,6,8,10,12,14,16,18}; int b[9]={3,1,5,9,4,2,7,6,10}; int t[2]; vector<string> splitEx(const string& src, string separate_chara

2018-08-02

2.5K0

静态数据竞争检测工具之LOCKSMITH-安装和使用

静态数据竞争检测工具都是基于中间语言开发的，在安装之前必须要先安装Ocaml和CIL

2018-08-02

5370

Redis源码阅读之rdb.c

REDIS_RDB_ENC_LZF |compressed_len | original_len | compressed_string

2018-08-02

6430

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