代码搜索：while
找到约 10,000 项符合「while」的源代码
代码结果 10,000
www.eeworm.com/read/440417/7689624
cpp gui.cpp
#include
void twomerge(int a[],int r[],int s,int m,int t)
{
int i,j,k;
i=s;j=m+1;k=s;
while(i
www.eeworm.com/read/440404/7689832
m instfreq.m
function [fnormhat,t]=instfreq(x,t,L,trace);
%INSTFREQ Instantaneous frequency estimation.
% [FNORMHAT,T]=INSTFREQ(X,T,L,TRACE) computes the instantaneous
% frequency of the analytic signal X at time
www.eeworm.com/read/440368/7690168
cpp 最小费用最大流.cpp
/***************************************************
*
* Ek形式的最小费用最大流矩阵形式。
* (poj 2516 Minimum Cost)
***************************************************/
int const MAXN = 256;//矩阵维数最大值
www.eeworm.com/read/440097/7693983
h app.h
/***************************************************************************
app.h - description
-------------------
begin
www.eeworm.com/read/440018/7695850
c msp430x47xx_uscib0_spi_01.c
//******************************************************************************
// MSP430x47xx Demo - USCI_B0, SPI Interface to TLC549 8-Bit ADC
//
// Description: This program demonstrate USC
www.eeworm.com/read/440002/7696024
c msp430xg46x_uscib0_spi_01.c
//******************************************************************************
// MSP430xG46x Demo - USCI_B0, SPI Interface to TLC549 8-Bit ADC
//
// Description: This program demonstrate USC
www.eeworm.com/read/439700/7702655
m aeldist.m
function x = AELDist(x0,m,beta,n)
format long;
x = zeros(n,1);
u = zeros(m,1);
for i=1:n
for j=1:m
r = MixMOD(x0,10,1);
k = 0;
while r(10) == 0
k = k +
www.eeworm.com/read/439700/7702657
m powerdist.m
function x = PowerDist(x0,beta,n)
x = zeros(n,1);
for i=1:n
r = MixMOD(x0,2,1);
k = 0;
while r(2) == 0
k = k + 1;
r(2) = power(2,k);
r = MixMOD(r(2),2,1);
www.eeworm.com/read/439700/7702690
m betap.m
function bp = betap(x,a,b)
%自变量的值：x
%第一个参数：a
%第二个参数：b
%自变量取x值时的不完全贝塔函数值：gp
format long;
if a
www.eeworm.com/read/439578/7705919
cpp game.cpp
#include
int comp( int x, int y)
{
if (x>y)
return 1;
else return 0;
}
int game( int group[], int n)
{
int j,i=n;
while (i>1)
{
i=i/2;
for (j=0;j