代码搜索：printf
找到约 10,000 项符合「printf」的源代码
代码结果 10,000
www.eeworm.com/read/406240/11446567
cpp zju1021.cpp
#include
int main()
{
int a[51]={0};
int output[51]={0};
int n;
int i,j;
char symbol;
int num=0;
scanf("%d",&n);
while(n!=0)
{
fflush(stdin);
scanf("%c",&symbol);
www.eeworm.com/read/405817/11456372
cpp sum it up(dfs生成不重复组合).cpp
#include
int sum,n,len,m;
int num[20],pre[20];
bool flag;
void print()
{
flag=true;
int i;
for(i=0; i
www.eeworm.com/read/405283/11466879
c 傻瓜递归.c
#include
main()
{ int m=1,n=1,s;
s=akm(m,n);
printf("%d",s);
}
akm(int m,int n)
{ if(m==0)
return n+1;
else if(m!=0&&n==0)
akm(m-1,1);
else if(m!=0&&n!=0)
www.eeworm.com/read/405283/11466930
c 小白鼠钻迷宫.c
这是个经典的数学问题，说的是：在一个随机的迷宫里，小白鼠如何能最快地从起点走到终点。
-# --------### # ## # # ###
----# # #-## # # # #### #### #
## ## #-# ## # # # #---# ## #
## # # # # -# ### ##
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c 小写数字转换成大写数字3.c
#include
void main()
{
double x,y;
char *ch[]={"零","壹","贰","叁","肆","伍","陆","柒","捌","玖"};
char *ch1[]={"拾","佰","仟","万","拾","佰","仟","亿"};
char num[256];
long i,n,j,m,y1;
printf("input:"
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c 逆阵.c
#include "stdio.h"
float z[4][4],*y=z; /*定义一个全局二维数组用来存放N-1阶余子式,因为A的伴随矩阵除以|A|时会产生小数，因此定义成float而非int*/
int js(int *p,int n) /*计算行列式的函数*/
{int k=0,i,s2=0,s1=0,j,s,t;
printf("\n"
www.eeworm.com/read/405283/11466938
c 逆矩阵.c
#define N 5 /*[注]：修改6为你所要的矩阵阶数*/
#include "stdio.h"
#include "conio.h"
/*js()函数用于计算行列式,通过递归算法实现*/
int js(s,n)
int s[][N],n;
{int z,j,k,r,total=0;
int b[N][N];/*b[N][N]用于存放，在矩阵s[
www.eeworm.com/read/405283/11466954
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/405283/11466957
c 猴子和桃.c
#include
main()
{
int x,x1,x2,x3,x4;/*我用的不知算不算是穷举法，桃子的初值设的是１００*/
for(x=100;x
www.eeworm.com/read/405283/11466966
c re.c
#include
main()
{
int n,s,j,k,p;
int a[100];//最多允许100人
printf("请输入人数：");
scanf("%d",&n);
printf("请输入数字：");
scanf("%d",&s);
for(j=0;j