题目描述如下: The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility) P A H N A P L S I I G Y I R And then read line by line: "PA…

题目描述: The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility) P A H N A P L S I I G Y I R And then read line by line: "PAHN…

请用C语言实现将double类型数据转换成字符串,再转换成double类型的数据.int类型的数据 想要完成题目中的功能,首先我们的先对系统存储double的格式有所了解. 浮点数编码转换使用的是IEEE的编码标准. 类型 存储位数 ​ ​ ​ 偏置值 ​ ​ 数符(S) 阶码(E) 尾数(M) 总位数 十六进制 十进制 短浮点数(Single,float) 1位 8位 23位 32位 7FH +127 长浮点数(Double) 1位 11位 52位 64位 3FFH +1023 临时浮点数(扩…

ZigZag Conversion The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility) P A H N A P L S I I G Y I R And then read line by line…

Problem: The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility) P A H N A P L S I I G Y I R And then read line by line: "P…

这道题真的不难吧. 如@AKEE@AKEE@AKEE 大佬所说,此题的确可以将n推广到一般情况. 但题面还是良心的只到了N<=4N<=4N<=4 以目前的题目来看,简单模拟即可. 分别弄N=1,2,3,4N=1,2,3,4N=1,2,3,4的情况判断就行. 下面奉上std 代码: #include<bits/stdc++.h> #define L 1000001 using namespace std; long long int a[100001],b,c,d,e,x=0,…

[热烈庆祝ZOJ回归] P1002:简单的DFS #include <cstdio> #include <cstring> #include <algorithm> ][]; int N; int input() { scanf("%d",&N); ;i<N;i++) scanf("%s",map[i]); return N; } ][]; ,sizeof(disable)); } int dfs(int cur)…

#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; #define inf 10000000 #define MAXN 20000 struct Edge{ int x,y,len; }e[3*MAXN]; int n,m,r,ans; int fa[MAXN]; int find(int x) { if(x != fa[…

(最开始源于牛客网上的一道编程题 : [编程题] 数制转换) 将X进制转为int十进制的功能函数:(乘X次方各位数的加和法) # 将X进制转为int十进制的功能函数:(乘X次方各位数的加和法) def XToInt(strNum,X): sum = 0 strNum = strNum.lstrip(") #去掉开始的0(前导零) length = len(strNum) for i in range(length): sum += ABCToNum(strNum[i])*(X**(length-…

由于某个需求,需要如题的转换,废话不多说,直接贴代码了,其实挺丑了,备用了 func (this *Server) socketParserHandler(client *genTcpServer.Client, fullData []byte) { fmt.Println("original data is", client) strPointerHex := fmt.Sprintf("%p", unsafe.Pointer(client)) fmt.Printl…