代码搜索：continue
找到约 10,000 项符合「continue」的源代码
代码结果 10,000
www.eeworm.com/read/147693/12538675
pl fig12_3.pl
% Figure 12.3 A best-first search program.
% bestfirst( Start, Solution): Solution is a path from Start to a goal
bestfirst( Start, Solution) :-
expand( [], l( Start, 0/0), 9999, _, yes,
www.eeworm.com/read/147186/12578831
m montagekpm3.m
function montageKPM3(data)
% data{f}(y,x,b) - each frame can have a different size (can can even be empty)
Nframes = length(data);
Nbands = -inf;
nr = -inf; nc = -inf;
for f=1:Nframes
if isempty(da
www.eeworm.com/read/146860/12606973
m divider.m
function [N,M]=divider(N1);
%DIVIDER Find dividers of an integer.
% [N,M]=DIVIDER(N1) find two integers N and M such that M*N=N1 and
% M and N as close as possible from sqrt(N1).
%
% Example :
% N1
www.eeworm.com/read/334136/12631912
m split_test.m
function v=split_test(b,mindim,fun)
k=size(b,3)
v(1:k)=false;
for i=1:k
quadregion=b(:,:,i);
if size(quadregion,1)
www.eeworm.com/read/300587/13904422
cpp 1197 specialized four-digit numbers.cpp
/*
1197 Specialized Four-Digit Numbers
Time Limit : 1000 ms Memory Limit : 32768 K Output Limit : 256 K
GUN C++
*/
#include
using namespace std;
int main()
{
int t,n,i;
www.eeworm.com/read/134785/13973726
cpp simcat v2.cpp
#include
#include
#include
using namespace std;
inline void display(istream& istrm);
int main(int argc, char* argv[])
{
if (argc==1) display(cin);
f
www.eeworm.com/read/237026/13980812
m divider.m
function [N,M]=divider(N1);
%DIVIDER Find dividers of an integer.
% [N,M]=DIVIDER(N1) find two integers N and M such that M*N=N1 and
% M and N as close as possible from sqrt(N1).
%
% Example :
% N1
www.eeworm.com/read/236321/14021673
c flash.c
#include "msp430x14x.h"
#include "def.h"
//程序接口开始
#define SDAH P5DIR|=BIT6;P5OUT|=BIT6;_NOP()
#define SDAL P5DIR|=BIT6;P5OUT&=(~BIT6);_NOP()
#define SDAIN P5DIR&=(~BIT6)
#define SDApor
www.eeworm.com/read/133638/14032712
org divider.org
function [N,M]=divider(N1);
%DIVIDER Find dividers of an integer.
% [N,M]=DIVIDER(N1) find two integers N and M such that M*N=N1 and
% M and N as close as possible from sqrt(N1).
%
% Example :
% N1