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/371050/9570137
install-sh
#!/bin/sh
#
# install - install a program, script, or datafile
#
# This originates from X11R5 (mit/util/scripts/install.sh), which was
# later released in X11R6 (xc/config/util/install.sh) with the
#
www.eeworm.com/read/174714/9575897
c buzzard.c
#include
#define X(s) (!(s&3)-((s&3)==2))
#define W while
char Z[82][82],A,B,f,g=26;z(q){return atoi(q);}m(d,l){return
Z[ B + X ( f +
3) * d+l *X(f+ 2 )][ A+X ( f ) * d
www.eeworm.com/read/366666/9805441
cpp usaco_crypt1.cpp
/*
ID:wangyuc2
PROG:crypt1
LANG:C++
*/
#include
#include
#include
using namespace std;
ifstream fin("crypt1.in");
ofstream fout("crypt1.out");
int p[9],nu
www.eeworm.com/read/366666/9805450
cpp usaco_5_1_3_theme.cpp
/*
PROB: theme
LANG: C++
*/
/*
枚举题，它的枚举策略就很重要……
这道题，我先把这个序列相邻两个的差值算出来，然后再直接找差值中相同的，那么就不用再找那个升调了！
这道题，刚开始我3重循环的最外层是长度，然后从最长的往下搜，搜到就直接输出，结果第9组就卡住了。
后来想想，这样的话可能长度只有26，但是我却从2500搜，效率太低，所以就换成从短的往长的搜 ...
www.eeworm.com/read/366666/9805459
cpp usaco_3_2_4_ratios.cpp
/*
ID: wangyuc2
PROB:ratios
LANG: C++
*/
/*
枚举啦，100^3，范围很小。
不过数据是非负的，所以包含0，而0的处理很恶心！！
为了这个零的处理，我提交了4次才过~~
*/
#include
#include
#include
#include
#i
www.eeworm.com/read/366616/9806375
install-sh
#!/bin/sh
#
# install - install a program, script, or datafile
# This comes from X11R5 (mit/util/scripts/install.sh).
#
# Copyright 1991 by the Massachusetts Institute of Technology
#
# Permission to
www.eeworm.com/read/366581/9807800
in znew.in
:
#!/bin/sh
PATH="BINDIR:$PATH"; export PATH
check=0
pipe=0
opt=
files=
keep=0
res=0
old=0
new=0
block=1024
# block is the disk block size (best guess, need not be exact)
warn="(does not preserve mo
www.eeworm.com/read/366581/9807815
in gzexe.in
:
#!/bin/sh
# gzexe: compressor for Unix executables.
# Use this only for binaries that you do not use frequently.
#
# The compressed version is a shell script which decompresses itself after
# skippi
www.eeworm.com/read/170114/9818770
sh caseexample.sh
#!/bin/sh
echo –n "Do you want to continue this operation? [n]"
read yesno
case $yesno in
y | Y | Yes | yes)
echo "system will continue this operation"
;;
n | N | no | NO)
echo "system will