代码搜索：while
找到约 10,000 项符合「while」的源代码
代码结果 10,000
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txt 经典c程序100例81--90.txt
【程序81】
题目：809*??=800*??+9*??+1 其中??代表的两位数,8*??的结果为两位数，9*??的结果为3位数。求??代表的两位数，及809*??后的结果。
1.程序分析：
2.程序源代码：
output(long b,long i)
{ printf("\n%ld/%ld=809*%ld+%ld",b,i,i,b%i);
}
main()
{long int
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c 0_to_100.c
#include
void main ()
{
int value = 0;
while (value
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c no_cont.c
#include
void main()
{
int counter;
printf("\nEven values\n");
for (counter = 1; counter
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c chrcnt.c
int chrcnt(const char *string, int letter)
{
int count = 0;
while (*string)
if (*string == letter)
count++;
return(count);
}
www.eeworm.com/read/263754/11344620
c led.c
#include
int main(void)
{
unsigned char i,j,k,led=0;
DDRA = 0xff;
PORTA = 0xff;
while(1)
{
if(led)
PORTA |= 0x01;//黄灯灭
else
PORTA &= 0xfe;//黄灯亮
led = !led;
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c stdlib.c
/*
* File: stdlib.c
* Purpose: Functions normally found in a standard C lib.
*
* Notes: This supports ASCII only!!!
*
*/
#include "src/common/common.h"
#include "src/common/stdlib.h
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digsilent2psat
#!/usr/bin/perl -w
# DIGSILENT2PSAT converts DIGSILENT data file into PSAT data file
#
# DIGSILENT2PSAT FILEINPUT
#
# Author: Juan Carlos Morataya
# Date: 14-Oct-2005
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ge2psat
#!/usr/bin/perl -w
# GE2PSAT converts GE data file into PSAT data file
#
# GE2PSAT FILEINPUT
#
# Author: Juan Carlos Morataya
# Date: 25-Oct-2005
# Version: 1.0.0
#
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m sor.m
%Successive Over Relaxation迭代程序
function [x,sp]=sor(a,b,n,x1,w)
%误差
e=ones(n,1);
%迭代的解向量
x2=zeros(n,1);
%迭代的次数
k=0;
%当误差没有满足要求时继续迭代
while norm(e,2)>1e-6
%每隔5步显示迭代结果
if (rem(k,5)==0)
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m biology.m
clear;
n=10000;
x=zeros(n,1);
x(1)=0.1;
i=1;
while i