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[Windows Develop] stepnc_write_v1_1 DL : 1: The STEPNCWrite library writes STEP-NC toolpath data (AP-238 CC1). This C++ library is completely open-source and has been released under the GNU General Public License (GPL). The library writes data in XML format (ISO 10303-28) using just printf(), so it does not depend on any STEP toolkit. It has no other dependencies and should build on Windows, Linux, Mac, or any other Unix platform. If you want to view your toolpath files, you can use the open-source Java AP-238 Toolpath Viewer. For more sophisticated applications, we encourage you to look at our commercial product, the STEP-NC Explorer which can display workpiece, fixture, and tool geometry as well as the paths, simulate their operation and convert -STEPNCWrite writes STEP-NC to olpath data (AP-238 CC1). This C library is comp letely open-source and has been released under the GNU General Public License (GPL). The libra ry writes data in XML format (ISO 10303-28) usin g just printf (), so it does not depend on any STEP toolkit. It has n o other dependencies and should build on Window s, Linux, Mac, or any other Unix platform. If you want to view yo ur toolpath files, you can use the open-source Java AP-238 Toolpat h Viewer. For more sophisticated applications , we encourage you to look at our commercial pro duct , the STEP-NC Explorer which can display workp iece, fixture , and tool geometry as well as the paths , simulate their operation and convert
Date : 2008-10-13 Size : 98.37kb User : 易力
[Windows Develop] xitong DL : 0: 学生统计成绩，插入，删除等，char c do { system(\"cls\") /*运行前清屏*/ printf(\"****************************主菜单****************************\\n\") /*菜单选择*/ printf(\"\\t\\t 1. 输入存储学生信息 \\n\") printf(\"\\t\\t 2. 查看学生成绩 \\n\") printf(\"\\t\\t 3. 增加学生信息\\n\") printf(\"\\t\\t 4. 删除学生信息\\n\") printf(\"\\t\\t 5. 查找学生信息\\n\") printf(\"\\t\\t 6. 排序学生信息 \\n\") printf(\"\\t\\t 7. 修改学生信息 \\n\") printf(\"\\t\\t 8. 查看学生成绩统计最高分，最低分，平均值 \\n\") printf(\"\\t\\t 0. 退出系统 \\n\")
Date : 2008-10-13 Size : 2.45kb User : 伍国
[Windows Develop] formulaC++20080625 DL : 0: 该组件可用编写字符串和数字运算处理脚本，供运行时调用获取执行结果 1、组件调用的方式可以参考 main.cpp 2、脚本的编写可以参考 debug/my.cfg 3、可调用系统函数有 atof: 将字符串转换为浮点数 ceil: floor: abs: index: strcmp: substr: strlen: trim: strstr: split: replace:字符串替换函数 printf: 这些函数基本与同名c函数相同 4、可预先编写一组脚本文件，编译时为每个文件指定ID，供调用时选择
Date : 2008-10-13 Size : 424.52kb User : 范哲伟
[Windows Develop] ADDR DL : 0: 本源码主要应用PRINTF函数和程序地址的访问,属于C方面的高级编程
Date : 2008-10-13 Size : 18.58kb User : walky
[Windows Develop] 征服4.0马端源码 DL : 1: using System; using System.Collections.Generic; using System.Text; namespace NewestCOServer { public class Cryption { class CryptCounter { UInt16 m_Counter = 0; public byte Key2 { get { return (byte)(m_Counter >> 8); } } public byte Key1 { get { return (byte)(m_Counter & 0xFF); } } public void Increment() { m_Counter++; } } private CryptCounter _decryptCounter; private CryptCounter _encryptCounter; private byte[] _cryptKey1; private byte[] _cryptKey2; private byte[] _cryptKey3; private byte[] _cryptKey4; private bool Decrypt2 = false; public Cryption() { _decryptCounter = new CryptCounter(); _encryptCounter = new CryptCounter(); _cryptKey1 = new byte[0x100]; _cryptKey2 = new byte[0x100]; byte i_key1 = 0x9D; byte i_key2 = 0x62; for (int i = 0; i 4 | buffer[i] < 4 | buffer[i] < 4 | buffer[i] << 4); buffer[i] ^= (byte)(_cryptKey4[_decryptCounter.Key2] ^ _cryptKey3[_decryptCounter.Key1]); _decryptCounter.Increment(); } } } public void GenerateKeys(UInt32 CryptoKey, UInt32 AccountID) { UInt32 tmpkey1 = 0, tmpkey2 = 0; tmpkey1 = ((CryptoKey + AccountID) ^ (0x4321)) ^ CryptoKey; tmpkey2 = tmpkey1 * tmpkey1; _cryptKey3 = new byte[256]; _cryptKey4 = new byte[256]; for (int i = 0; i < 256; i++) { int right = ((3 - (i % 4)) * 8); int left = ((i % 4)) * 8 + right; _cryptKey3[i] = (byte)(_cryptKey1[i] ^ tmpkey1 left); _cryptKey4[i] = (byte)(_cryptKey2[i] ^ tmpkey2 left); } Decrypt2 = true; } public void GenerateKeys2(byte[] InKey1, byte[] InKey2) { byte[] addKey1 = new byte[4]; byte[] addKey2 = new byte[4]; byte[] addResult = new byte[4]; //addKey1.i = 0; //addKey2.i = 0; byte[] tempKey = new byte[4]; long LMULer; // InKey1[0] = 0x20; // InKey1[1] = 0x5c; // InKey1[2] = 0x48; // InKey1[3] = 0xf4; // InKey2[0] = 0x00; // InKey2[1] = 0x44; // InKey2[2] = 0xa6; // InKey2[3] = 0x2e; //if (Key3) delete [] Key3; //if (Key4) delete [] Key4; _cryptKey3 = new byte[256]; _cryptKey4 = new byte[256]; for (int x = 0; x < 4; x++) { addKey1[x] = InKey1[3 - x]; addKey2[x] = InKey2[3 - x]; } //cout << "Key1: " << addKey1.i << endl; //cout << "Key2: " << addKey2.i << endl; uint Adder1; uint Adder2; uint Adder3; Adder1 = (uint)((addKey1[3] << 24) | (addKey1[2] << 16) | (addKey1[1] << 8) | (addKey1[0])); Adder2 = (uint)((addKey2[3] << 24) | (addKey2[2] << 16) | (addKey2[1] 8) & 0xff); addResult[2] = (byte)((Adder3 >> 16) & 0xff); addResult[3] = (byte)((Adder3 >> 24) & 0xff); for (int b = 3; b >= 0; b--) { // printf("%.2x ", addResult.c[b]); tempKey[3 - b] = addResult[b]; } tempKey[2] = (byte)(tempKey[2] ^ (byte)0x43); tempKey[3] = (byte)(tempKey[3] ^ (byte)0x21); for (int b = 0; b < 4; b++) { tempKey[b] = (byte)(tempKey[b] ^ InKey1[b]); } //Build the 3rd Key for (int b = 0; b < 256; b++) { _cryptKey3[b] = (byte)(tempKey[3 - (b % 4)] ^ _cryptKey1[b]); } for (int x = 0; x < 4; x++) { addResult[x] = tempKey[3 - x]; } Adder3 = (uint)((addResult[3] << 24) | (addResult[2] << 16) | (addResult[1] << 8) | (addResult[0])); LMULer = Adder3 * Adder3; LMULer = LMULer 32; Adder3 = Convert.ToUInt32(LMULer & 0xffffffff); addResult[0] = (byte)(Adder3 & 0xff); addResult[1] = (byte)((Adder3 >> 8) & 0xff); addResult[2] = (byte)((Adder3 >> 16) & 0xff); addResult[3] = (byte)((Adder3 >> 24) & 0xff); for (int b = 3; b >= 0; b--) { tempKey[3 - b] = addResult[b]; } //Build the 4th Key for (int b = 0; b < 256; b++) { _cryptKey4[b] = Convert.ToByte(tempKey[3 - (b % 4)] ^ _cryptKey2[b]); } Decrypt2 = true; //cout << "Int representation: " << charadd.i << endl; } } }
Date : 2010-04-08 Size : 3.47mb User : andesion@vip.qq.com
[Windows Develop] stepnc_write_v1_1 DL : 0: The STEPNCWrite library writes STEP-NC toolpath data (AP-238 CC1). This C++ library is completely open-source and has been released under the GNU General Public License (GPL). The library writes data in XML format (ISO 10303-28) using just printf(), so it does not depend on any STEP toolkit. It has no other dependencies and should build on Windows, Linux, Mac, or any other Unix platform. If you want to view your toolpath files, you can use the open-source Java AP-238 Toolpath Viewer. For more sophisticated applications, we encourage you to look at our commercial product, the STEP-NC Explorer which can display workpiece, fixture, and tool geometry as well as the paths, simulate their operation and convert -STEPNCWrite writes STEP-NC to olpath data (AP-238 CC1). This C library is comp letely open-source and has been released under the GNU General Public License (GPL). The libra ry writes data in XML format (ISO 10303-28) usin g just printf (), so it does not depend on any STEP toolkit. It has n o other dependencies and should build on Window s, Linux, Mac, or any other Unix platform. If you want to view yo ur toolpath files, you can use the open-source Java AP-238 Toolpat h Viewer. For more sophisticated applications , we encourage you to look at our commercial pro duct , the STEP-NC Explorer which can display workp iece, fixture , and tool geometry as well as the paths , simulate their operation and convert
Date : 2025-12-21 Size : 98kb User : 易力
[Windows Develop] xitong DL : 0: 学生统计成绩，插入，删除等，char c do { system("cls") /*运行前清屏*/ printf("****************************主菜单****************************\n") /*菜单选择*/ printf("\t\t 1. 输入存储学生信息 \n") printf("\t\t 2. 查看学生成绩 \n") printf("\t\t 3. 增加学生信息\n") printf("\t\t 4. 删除学生信息\n") printf("\t\t 5. 查找学生信息\n") printf("\t\t 6. 排序学生信息 \n") printf("\t\t 7. 修改学生信息 \n") printf("\t\t 8. 查看学生成绩统计最高分，最低分，平均值 \n") printf("\t\t 0. 退出系统 \n")-Student achievement statistics, insert, delete, etc., char c do (system (cls)/* run Qianqing screen*/printf (*********************** Main Menu*********************************)/* menu*/printf (1. input storage Student Information ) printf (2. See student achievement) printf (3. increasing the number of students of information) printf (4. delete student information) printf (5. Student Information search) printf (6. sort student information) printf (7. modify student information) printf (8. See the highest score student performance statistics, the lowest points average) printf (0. Exit System)
Date : 2025-12-21 Size : 2kb User : 伍国
[Windows Develop] tony DL : 0: #include <stdio.h> #include <stdlib.h> enum x {A,B,C,D,E} int main(void) { int a=73,b=85,c=66 { if (a>=90) printf("a=A等級!!\n") else if (a>=80) printf("73分=B等級!!\n") else if (a>=70) printf("73分=C等級!!\n") else if (a>=60) printf("73分=D等級!!\n") else if (a<60) printf("73分=E等級!!\n") } { if (b>=90) printf("b=A等級!!\n") else if (b>=80) printf("85分=B等級!!\n") else if (b>=70) printf("85分=C等級!!\n") else if (b>=60) printf("85分=D等級!!\n") else if (b<60) printf("85分=E等級!!\n") } { if (c>=90) printf("c=A等級!!\n") else if (c>=80) printf("66分=B等級!!\n") else if (c>=70) printf("66分=C等級!!\n") else if (c>=60) printf("66分=D等級!!\n") else if (c<60) printf("66分=E等級!!\n") } system("pause") return 0 }-# Include <stdio.h># Include <stdlib.h> enum x (A, B, C, D, E) int main (void) (int a = 73, b = 85, c = 66 (if ( a> = 90) printf ( a = A grades!! ) else if (a> = 80) printf ( 73 points = B grade!! ) else if (a> = 70) printf ( 73 points = C grade!! ) else if (a> = 60) printf ( 73 points = D grade!! ) else if (a <60) printf ( 73 points = E grades!! )) (if (b > = 90) printf ( b = A grades!! ) else if (b> = 80) printf ( 85 points = B grade!! ) else if (b> = 70) printf ( 85 points = C grade!! ) else if (b> = 60) printf ( 85 points = D grade!! ) else if (b <60) printf ( 85 points = E grades!! )) (if (c> = 90) printf ( c = A grades!! ) else if (c> = 80) printf ( 66 points = B grade!! ) else if (c> = 70) printf ( 66 points = C grades !! ) else if (c> = 60) printf ( 66 points = D grade!! ) else if (c <60) printf ( 66 points = E grades!! )) system ( pause ) return 0)
Date : 2025-12-21 Size : 31kb User : 林小世
[Windows Develop] Ar1 DL : 0: 在c语言中打开一个英文文档，用Si（i<=文章句子数）代替文章中的各句子。 - {while(ch=getchar()) { if ((ch>= a &&ch<= z )||(ch>= A &&ch<= Z )) {ch=ch+7 if(ch> Z &&ch<= Z +7||ch> z ) ch=ch-26 printf(" c",ch) } else if(ch!= \n )
Date : 2025-12-21 Size : 2kb User : bylon
[Windows Develop] Greatlist DL : 0: 声明并赋初始值。j,k作为循环计数变量仅声明不赋初始值。第一个循环j从1到4，控制打印行数；内部循环从0到2*j-1(对应j分别取值1,3,5,7)，这个循环控制打印字符的个数；printf(" c",i) 这一句把i变量按相应的ASCII码对应的字符以字符形式打印出来；-Statement and to give the initial value. j, k as the loop counter not only to declare the initial value assigned. The first cycle j from 1 to 4, control print lines internal loop from 0 to 2* j-1 (corresponding to j values 1,3,5,7, respectively), the loop control the number of printed characters printf ( " c" , i) this one the i variable by the corresponding ASCII character code corresponding to the characters to be printed
Date : 2025-12-21 Size : 1kb User : Gigi
[Windows Develop] Printf DL : 0: 该源码的主要功能：用C来实现带有缺省参数的函数，风格规范，功能齐全，能深入理解指针的强大性以及函数printf神奇的背后。-The main function of the source: Using C to implement a function with default parameters, style specifications, functional and powerful pointer to in-depth understanding of and behind the magic function printf.
Date : 2025-12-21 Size : 10kb User : 李晨旭
[Windows Develop] C DL : 0: printf("Please shuru zhengzhiscore:") gets(temp) data.score[3]=atof(temp) data.score[4]=data.score[0]+data.score[1]+data.score -gets(temp) data.score[1]=atof(temp) printf("Please input yingyu score:") gets(temp) data.score[2]=atof(temp) printf("Please shuru zhengzhiscore:") gets(temp) data.score[3]=atof(temp)
Date : 2025-12-21 Size : 5kb User : wangming
[Windows Develop] classical-algorithm-c-language DL : 0: c语言经典算法【程序1】题目：有1、2、3、4个数字，能组成多少个互不相同且无重复数字的三位数？都是多少？ 1.程序分析：可填在百位、十位、个位的数字都是1、2、3、4。组成所有的排列后再去掉不满足条件的排列。 2.程序源代码： main() { int i,j,k printf("\n") for(i=1 i<5 i++) ／*以下为三重循环*/ for(j=1 j<5 j++) for (k=1 k<5 k++) { if (i!=k&&i!=j&&j!=k) /*确保i、j、k三位互不相同*/ printf(" d, d, d\n",i,j,k) } } -classical algorithm c language program 1】【Title: There are numbers 1,2,3,4, how many can be composed of distinct three-digit numbers with no repetition? Is how much? 1. Program Analysis: fill in the hundred, ten, a bit of the numbers are 1,2,3,4. Removed after the composition does not satisfy all the conditions of the arrangement of the arrangement. 2. Program source code: main () {int i, j, k printf (" \ n" ) for (i = 1 i < 5 i++) /* The following is the triple loop*/for (j = 1 j < 5 j++) for (k = 1 k < 5 k++) {if (i! = k & & i! = j & & j! = k) /* ensure i, j, k three distinct*/printf ( " d, d, d \ n" , i, j, k) }}
Date : 2025-12-21 Size : 69kb User : zhang
[Windows Develop] printf DL : 0: 《轻松学通Visual C++》的源代码--打字测试系-" Easy to learn through the Visual C++" source code- typing test system
Date : 2025-12-21 Size : 327kb User : 天涯
[Windows Develop] summary-of-c-language-printf-form DL : 0: c/c++输出格式设置总结，希望对大家有所帮助-c/c++ output format summary
Date : 2025-12-21 Size : 15kb User : xf
[Windows Develop] C-Programming-Language-100-cases DL : 0: 文件包含100个C语言程序的例子,由易到难,可以帮助初学者提高学习效率。注:1、2、3、4个数字,有多少能形成一个不同于彼此,没有重复数字三位数吗?是多少? 1。这个程序分析:能填补一百年,十,比特数1、2、3、4。所有的安排后再删除不满足条件的安排。 2.程序源码为： main() { int i,j,k printf("\n") for(i=1 i<5 i++) for(j=1 j<5 j++) for (k=1 k<5 k++) { if (i!=k&&i!=j&&j!=k) printf(" d, d, d\n",i,j,k) } } -File consists of 100 C program example, from easy to difficult, can help beginners improve the learning efficiency.Topic: 1, 2, 3, 4 Numbers, how many can form a different from each other and have no the repeat number three digits? How much be?1. The program analysis: can fill in one hundred, ten, bit number is 1, 2, 3, 4. Of all the arrangement of remove not meet again after the arrangement of the condition. 2. The source code： main() { int i,j,k printf("\n") for(i=1 i<5 i++) for(j=1 j<5 j++) for (k=1 k<5 k++) { if (i!=k&&i!=j&&j!=k) printf(" d, d, d\n",i,j,k) } } ……
Date : 2025-12-21 Size : 6.55mb User : liz
[Windows Develop] summary-of-c-language-printf-form DL : 0: c/c++输出格式设置总结，希望对大家有所帮助-c/c++ output format summary
Date : 2025-12-21 Size : 15kb User : hecoun
[Windows Develop] printf DL : 0: 《轻松学通Visual C++》的源代码--打字测试系-" Easy to learn through the Visual C++" source code- typing test system
Date : 2025-12-21 Size : 327kb User : liuzhi
[Windows Develop] printf DL : 0: 《轻松学通Visual C++》的源代码--打字测试系-" Easy to learn through the Visual C++" source code- typing test system
Date : 2025-12-21 Size : 328kb User : erandto
[Windows Develop] 868123 DL : 0: #include <stdio.h> 　　#include <stdlib.h> 　　//输入十进制数N和转化的进制数M 　　void trans(int n,int m) 　　{ char str[100] int i 　　　for(i=0 n>0 i++) {if (n m<10) {str[i]=n m+ 0 } else {str[i]=n m-10+ A } n=n/m } for(n=i n>0 n--){printf(" c",str[n-1]) } 　　} 　　void main() 　{ int m,n,x char ch 　printf("给定进制 M---") 　scanf(" d",&m) 　loop: 　printf("给定一个 d 进制的数 X---",m) 　fflush(stdin) //一个M进制的数X转10进制　for(x=0 ) 　{ ch=getchar() 　if(ch>= 0 && ch<= 9 ) { n=ch- 0 } 　else if(ch>= a && ch<= z ) { n=ch- a +10 } 　else if(ch>= A && ch<= Z ) { n=ch- A +10 } 　else { break } 　if(n>=m){goto loop } 　x=x*m+n 　} 　printf("转换成 10 进制的数为 --- d\n",x) 　printf("给定要转换成的进制 N---") 　scanf(" d",&m) 　printf("转换成 d 进制后的结果 ---",m) 　trans(x,m) 　printf("\n") }-#include <stdio.h> 　　#include <stdlib.h> 　　//输入十进制数N和转化的进制数M 　　void trans(int n,int m) 　　{ char str[100] int i 　　　for(i=0 n>0 i++) {if (n m<10) {str[i]=n m+ 0 } else {str[i]=n m-10+ A } n=n/m } for(n=i n>0 n--){printf(" c",str[n-1]) } 　　} 　　void main() 　{ int m,n,x char ch 　printf("给定进制 M---") 　scanf(" d",&m) 　loop: 　printf("给定一个 d 进制的数 X---",m) 　fflush(stdin) //一个M进制的数X转10进制　for(x=0 ) 　{ ch=getchar() 　if(ch>= 0 && ch<= 9 ) { n=ch- 0 } 　else if(ch>= a && ch<= z ) { n=ch- a +10 } 　else if(ch>= A && ch<= Z ) { n=ch- A +10 } 　else { break } 　if(n>=m){goto loop } 　x=x*m+n 　} 　printf("转换成 10 进制的数为--- d\n",x) 　printf("给定要转换成的进制 N---") 　scanf(" d",&m) 　printf("转换成 d 进制后的结果---",m) 　trans(x,m) 　printf("\n") }
Date : 2025-12-21 Size : 9kb User : 868123

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