#include <stdio.h>
void daoxu(char str[]){
int i;
char temp;
for(i = 0; i < strlen(str) / 2 ; i ++){
temp = str[i];
str[i] = str[strlen(str) - i-1];
str[strlen(str) - i-1] = temp;
}
}
int wordCount(char str[]){
int i;
int count = 0;
int word = 0;
for (i = 0 ; str[i] != '\0' ; i ++)
{
if (str[i] == ' ')
{
word = 0;
}else if (word == 0)
{
word = 1;
count ++;
}
}
return count;
}
void mylwr(char str[]){
int i;
for (i = 0 ; str[i] != '\0' ; i ++)
{
if (str[i] >= 'A' && str[i] <= 'Z')
{
str[i] += 'a' - 'A';
}
}
}
void myupr(char str[]){
int i;
for (i = 0 ; str[i] != '\0' ; i ++)
{
if (str[i] >= 'a' && str[i] <= 'z')
{
str[i] -= 'a' - 'A';
}
}
}
int mylen(char str[]){
int len;
for (len = 0 ; str[len] != '\0' ; len ++);
return len;
}
void mycat(char str1[],char str2[]){
int i,j;
for (i = 0 ; str1[i] != '\0' ;i++);
for (j = 0 ; str2[j] != '\0' ; j ++)
{
str1[i + j] = str2[j];
}
str1[i + j] = '\0';
}
void myncat(char str1[],char str2[], int len){
int i,j;
for(i = 0; str1[i] != '\0'; i++);
for (j = 0; j < len; j++)
{
str1[i + j] = str2[j];
}
str1[i + j] = '\0';
}
void mycpy(char dst[],char src[]){
int i = 0;
do
{
dst[i] = src[i];
} while (src[i++] != '\0');
}
void myncpy(char dst[],char src[], int len){
int i;
for (i = 0; i < len; i++)
{
dst[i] = src[i];
}
dst[i] = '\0';
}
void longest(char dst[],char src[]){
int i = 0,j;
int count =0;
int max = -1;
do
{
if (src[i] ==' ' || src[i] == '\0')
{
if (count > max)
{
max = count;
for (j = 0; j < count; j++)
{
dst[j] = src[i - count + j];
}
dst[j] = '\0';
}
count = 0;
}else{
count++;
}
} while (src[i++] != '\0');
}
#include <stdio.h>
int getint(char str[], int a[]){//从字符串中提取数字并放在数组中
int i = 0;
int w = 0;
int c = 0;
int j, k;
do
{
if (str[i] > '0' && str[i] <= '9')
{
w++;
}else if (w)
{
j = 0;
for (k = w; k > 0; k--)
{
j *= 10;
j += str[i - k] - '0';
}
w = 0;
a[c] = j;
c++;
}
} while (str[i++] != '\0');
return c;
}
void main(){
char str[100];
int a[100];
int i, j;
gets(str);
i = getint(str,a);
for (j = 0; j < i; j++)
{
printf("%d ",a[j]);
}
}
#include <stdio.h>
#include <stdlib.h>
int sumof1(int x)//求一个数转换成二进制以后1的个数
{
int countx = 0;
while(x)
{
countx ++;
x &= x-1; //每位与一次x - 1;就能消掉最后一个1
}
return countx;
}
void main(){
char c[10];
int i = 999;
itoa(i, c, 10);//以10进制转换成字符数组
puts(c);
itoa(i, c, 16);//以16进制转换成字符数组
printf("0x%s\n", c);
itoa(i, c, 8);//以8进制转换成字符数组
printf("0%s\n", c);
itoa(i, c, 2);//以2进制转换成字符数组
puts(c);
i = atoi(c);//再将字符串转成整形
printf("%d\n", i);
printf("%d\n", sumof1(i));//以2进制表示时1的个数
}
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