代码搜索：integer
找到约 10,000 项符合「integer」的源代码
代码结果 10,000
www.eeworm.com/read/392695/8329830
txt test.txt
PROGRAM test ;
INTEGER : test1 ;
BEGIN
test1 := 2;
END.
www.eeworm.com/read/292735/8337201
bas pubdefines.bas
Attribute VB_Name = "PubDefine"
'PubDefine.bas
'公共定义，全局变量
'- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
'Stone State
Public Const NONE As Integer = 0
P
www.eeworm.com/read/292735/8337233
bas controlrcx.bas
Attribute VB_Name = "ControlRCX"
'ControlRCX.bas
'Contain some funcition that they could control RCX to do something.
'Some of them are interfaces
'- - - - - - - - - - - - - - - - - - - - - - -
www.eeworm.com/read/292622/8344929
bak p1098m.bak
{
思路非常简单：从左求一遍最长不连续递增子序列，再从右求一遍，二者相加(多加了一个本身）求最大值得max，max需要减1（减去一个本身）然后用n-max
最长不连续递增序列状态转移方程：ll[j]:=max(ll[j],ll[k]+1); 1
www.eeworm.com/read/292622/8344940
pas p1098.pas
program p1098;
var
i,j,n,max:integer;
t,up,down:array[1..100] of integer;
begin
readln(n);
for i:=1 to n do
read(t[i]);
up[1]:=1;
for i:=2 to n do begin
for j:=1 to i-1 do
www.eeworm.com/read/292622/8344962
bak p1098m.pas.bak
{
思路非常简单：从左求一遍最长不连续递增子序列，再从右求一遍，二者相加(多加了一个本身）求最大值得max，max需要减1（减去一个本身）然后用n-max
最长不连续递增序列状态转移方程：ll[j]:=max(ll[j],ll[k]+1); 1
www.eeworm.com/read/292622/8344976
pas p1098m.pas
{
思路非常简单：从左求一遍最长不连续递增子序列，再从右求一遍，二者相加(多加了一个本身）求最大值得max，max需要减1（减去一个本身）然后用n-max
最长不连续递增序列状态转移方程：ll[j]:=max(ll[j],ll[k]+1); 1
www.eeworm.com/read/292622/8345012
bak p1118(2).pas.bak
program p1118;
const
MAXS=6;
MAXK=40;
var
str,line:string;
word:array[1..MAXS] of string;
i,j,m,p,k,s:integer;
f:array[0..200,0..MAXK] of integer;
b:array[1..200,1..200] of integer;
www.eeworm.com/read/292622/8345060
pas p1311.pas
program dcjl;
const
MAXN=50;
var
map:array[1..MAXN,1..MAXN] of integer;
used:array[1..MAXN] of 0..2;
word:array[1..MAXN] of string;
n,i,max:integer;
t,ch:char;
f
www.eeworm.com/read/292622/8345101
pas p1340.pas
program p1340(input,output);
{$M 16777216}
label 99;
const
MAXN=30;
MAXM=30;
dx:array[1..4] of integer=(-1,1,0,0);
dy:array[1..4] of integer=(0,0,-1,1);
var
map:array[0..M