代码搜索：预置数
找到约 10,000 项符合「预置数」的源代码
代码结果 10,000
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txt 两数相加.txt
10 rem determine and print the sum of two integers
15 REM
20 rem input the two integers
30 input a
40 input B
45 rem
50 rem add integers and store result in c
60 let c = a + b
65 rem
70 rem p
www.eeworm.com/read/334029/12647032
cpp 数独算法.cpp
#include
#include
#include
enum{SIZE=81};
unsigned int Data[SIZE]={//未解棋盘数据
0 , 9 , 0 , 0 , 6 , 0 , 5 , 4 , 8 ,
4 , 0 , 3 , 0 , 8 , 0 , 9 , 0 , 0 ,
8 , 6
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cpp 1406 完数.cpp
/*
1406 完数
Time Limit : 1000 ms Memory Limit : 32768 K Output Limit : 256 K
15 MS 344 KB 874 B
GUN C++
*/
#include
#include
using namespace std;
const int Max=100
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cpp 1261 字串数.cpp
/*
1261 字串数
Time Limit : 1000 ms Memory Limit : 32768 K Output Limit : 256 K
GUN C++
*/
//组合公式很简单，组合运算量不大，只涉及大数乘除
#include
#include
#include
#include
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ppt 数的识别.ppt
www.eeworm.com/read/134640/13978715
txt 随机数.txt
01010100101100100000010101111010000000100010110001010000001001110000011001110100011111011000000000011010011100100000110110001001001001011110110101011001011110000001001110100011010001011110101000101111
www.eeworm.com/read/236748/14000188
txt 随机数.txt
#include "iostream.h"
#include "stdlib.h"
int main(int argc, char* argv[])
{
int num;
coutnum)
{
int b=0,f=0,i;
for(i=0;i
www.eeworm.com/read/133496/14038699
c 阿姆斯特朗数.c
#include
main()
{
int i,t,k,a[3];
printf("There are following Armstrong number smaller than 1000: \n");
for(i=152;i=10;t++)
{
a[t]=(i%k)/(k/1
www.eeworm.com/read/203529/15356479
cpp 优先数调度.cpp
#include
#define running 1 //用running表示进程处于运行态
#define aready 2 //用aready表示进程处于就绪态
#define blocking 3 //用blockin