1、从字符数组中读出相应的整数、实数。（写的有点可怕，先找第一个数字就会简单很多）
//从一个字符数组中读出相应的整数、实数
#include<stdio.h>
#include<math.h>
#include<string.h>
int main()
{
void atoif(char a[]);
char a[30];
fgets(a,30,stdin);
atoif(a);
return 0;
}
void atoif(char a[])
{
int i = 0;
int j = 0;
int l = strlen(a);
float b = 0;
int n = 1;
int c = 0;
while(a[i]!='\0'){
if(a[i]<='9'&&a[i]>='0'&&a[i-1]!='.'&&a[i+1]!='.'){
j = i;
while(a[j]<='9'&&a[j]>='0'){
c = c*10+(a[j]-'0');
j++;
}
if(a[j]!='.'){
printf("%d\n",c);
break;
}
}
if(a[i] == '.'){
if((i == 0||a[i-1]<='0'||a[i-1]>='9')&&(a[i+1]<='9'&&a[i+1]>='0')){
j = i+1;
while((a[j]<='9'&&a[j]>='0')&&j<l){
b = b + (a[j]-'0')*pow(0.1,n);
n++;
j++;
}
}
else if((i == l-1 ||a[i+1]<='0'||a[i+1]>='9')&&(a[i-1]<='9'&&a[i-1]>='0')){
j = i-1;
n = 0;
while((a[j]<='9'&&a[j]>='0')&&j>-1){
b+=(a[j]-'0')*pow(10,n);
n++;
j--;
}
}
else if((a[i-1]<='9'&&a[i-1]>='0')&&(a[i+1]<='9'&&a[i+1]>='0')){
j = i+1;
while((a[j]<='9'&&a[j]>='0')&&j<l){
b = b + (a[j]-'0')*pow(0.1,n);
n++;
j++;
}
j = i-1;
n = 0;
while((a[j]<='9'&&a[j]>='0')&&j>-1){
b+=(a[j]-'0')*pow(10,n);
n++;
j--;
}
}
printf("%f\n",b);
break;
}
//else
i++;
}
}
2、整数转换为一个字符数组
#include<stdio.h>
int main(){
void itoa(int a,char *b);
int a;
char b[10];
printf("please input a integer:");
scanf("%d",&a);
itoa(a,b);
printf("%s\n",b);
return 0;
}
void itoa(int a,char *b)
{
int i = 0;
int j = 0;
char t;
while(a!=0)
{
b[i] = (a%10)+'0';
a = a/10;
i++;
}
b[i] = '\0';
i = i-1;
while(j<i)
{
t = b[j];
b[j] = b[i];
b[i] = t;
j++;
i--;
}
}
3、字符串数组的冒泡排序（从小到大）（有个理解错误的问题，一开始输入的单词就应该用二维数组接着，不应该用一个大字符数组，还得提取出来放到二维数组里）
#include<stdio.h>//字符串数组的冒泡排序，从小到大
#include<string.h>
#define M 30
#define N 15
int main()
{
int bubble(char *a,char b[][M]);
char a[M];
char b [N][M];
int i;
int n;
printf("please input a string:");
scanf("%[^\n]",a);
n = bubble(a,b);
for(i = 0;i <= n;i++)
{
printf("%s\n",b[i]);
}
return 0;
}
int bubble(char *a,char b[][M])
{
int i = 0;
int j = 0;
int k = 0;
char c[M];
while(a[i]!='\0')
{
if(a[i]==' ')
{
b[j][k] = '\0';
j++;
k = 0;
}
else
{
b[j][k] = a[i];
k++;
}
i++;
}
for(i = 0;i <= j;i++)
{
for(k = i;k <= j;k++)
{
if(strcmp(b[i],b[k])>0)
{
strcpy(c,b[i]);
strcpy(b[i],b[k]);
strcpy(b[k],c);
}
}
}
return j;
}
4、素数，有且只有两个相同数的数
#include<stdio.h>
int main()
{
void primetcom(int a,int b);
int a,b;
printf("please input two number:");
scanf("%d%d",&a,&b);
primetcom(a,b);
return 0;
}
void primetcom(int a,int b)
{
int i;
int j;
int k;
int temp;
int count;
int n = 0;
int c[5];
for(;a <= b;a++){
for(i = 2;i < a;i++)
{
if(a%i==0)
{
break;
}
}
if(i == a)
{
temp = a;
j = 0;
count = 0;
while(temp != 0)
{
c[j] =temp%10;
k = 0;
while(k<j)
{
if(c[k]==c[j])
{
count++;
}
k++;
}
temp = temp/10;
j++;
}
if(count == 1){
printf("%d\n",a);
n++;
}
}
}
if(n == 0)
{
printf("not found a num fits your requirements!\n");
}
}
5、b字符串先返回来，和a穿插到c中
#include<stdio.h>
#include<string.h>
#define N 10
int main()
{
void change(char *a,char *b,char *c);
char a[N];
char b[N];
char c[2*N];
printf("please input two string:");
scanf("%s%s",a,b);
change(a,b,c);
printf("mix is %s\n",c);
return 0;
}
void change(char *a,char *b,char *c)
{
int i = 0;
int j;
int k;
char t;
j = strlen(b)-1;
while(i < j)
{
t = b[i];
b[i] = b[j];
b[j] = t;
i++;
j--;
}
printf("%s\n",b);
i = 0;
while(a[i/2]!='\0'&&b[i/2]!=0)
{
if(i%2)
{
c[i] = b[i/2];
}
else
{
c[i] = a[i/2];
}
i++;
}
j = i/2;
if(a[j]!='\0')
{
while(a[j]!='\0')
{
c[i] = a[j];
i++;
j++;
}
}
else if(b[j]!='\0')
{
while(b[j]!='\0')
{
c[i] = b[j];
i++;
j++;
}
}
c[i] = '\0';
}
6、二维数组，偶数行从小到大，奇数行从大到小
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#define N 5
int main()
{
void Output(int a[][N]);
void sort(int a[][N]);
int i;
int j;
int a[N][N];
srand((unsigned)time(NULL));
for(i = 0;i < N;i++)
{
for(j = 0;j < N;j++)
{
a[i][j] = rand()%10;
}
}
Output(a);
sort(a);
Output(a);
return 0;
}
void Output(int a[][N])
{
int i,j;
for(i = 0;i < N;i++)
{
for(j = 0;j < N;j++)
{
printf("%4d",a[i][j]);
}
printf("\n");
}
}
void sort(int a[][N])
{
int i;
int j;
int k;
int temp;
for(k = 0;k < N;k++)
{
if(k%2)
{
for(i = 0; i < N;i++)
{
for(j = i;j < N; j++)
{
if(a[k][i] < a[k][j])
{
temp = a[k][i];
a[k][i] = a[k][j];
a[k][j] = temp;
}
}
}
}
else
{
for(i = 0; i < N;i++)
{
for(j = i;j < N; j++)
{
if(a[k][i] > a[k][j])
{
temp = a[k][i];
a[k][i] = a[k][j];
a[k][j] = temp;
}
}
}
}
}
}
7、Troitsky（整数的第一位拿到最后一位，是否是原来数的整数倍）
#include<stdio.h>
#define N 100
int main()
{
int Troitsky(long a[]);
long a[N]={0};
int n;
int i = 0;
n = Troitsky(a);
printf("the total number is %d\n",n);
while(a[i] != 0)
{
printf("%ld\n",a[i]);
i++;
}
return 0;
}
int Troitsky(long a[])
{
int n = 0;
int i;
int b;
int j;
int k;
int num[5];
int newnum;
for(i = 10;i<=10000;i++)
{
b = i;
j = 0;
k = 0;
newnum = 0;
while(b != 0){
num[j] = b%10;
b = b/10;
j++;
}
for(k = j-2;k > -1;k--){
newnum = newnum*10 + num[k];
}
newnum = newnum*10 +num[j-1];
printf("%d\n",newnum);
if((newnum%i)==0)
{
n++;
a[n-1] = i;
}
}
return n;
}
8、将两个字符串合并成一个不重复单词的字符串
#include<stdio.h>
#include<string.h>
#define N 10
#define M 20
int main()
{
int i=0;
void set(char a[][M],char b[][M],char c[][M]);
char a[N][M]={{"while"},{"for"},{"switch"},{"if"},{"continue"}};
char b[N][M]={{"for"},{"case"},{"do"},{"else"},{"char"},{"switch"}};
char c[2*N][M]={'\0'};
set(a,b,c);
while(c[i][0]!='\0')
{
printf("%s\n",c[i]);
i++;
}
return 0;
}
void set(char a[][M],char b[][M],char c[][M]){
int i = 0;
int j = 0;
int k = 0;
while(b[i][0]!='\0')
{
strcpy(c[i],b[i]);
i++;
}
while(a[j][0]!='\0')
{
for(k = 0;k < i;k++)
{
if(strcmp(a[j],c[k])==0)
{
break;
}
}
if(k == i)
{
strcpy(c[i],a[j]);
i++;
}
j++;
}
}
9、找出7个莫尼森数，数是素数，2^数次方-1也是素数
#include<stdio.h>
#include<math.h>
int main()
{
int isprime(int n);
int i = 2;
int a;
int n;
int count = 0;
while(1)
{
n = isprime(i);
if(n == 1)
{
a = pow(2,i)-1;
n = isprime(a);
}
if(n == 1)
{
count++;
printf("%d is a Mersenynum!\n",i);
}
if(count == 7)
{
break;
}
i++;
}
printf("%d\n",count);
return 0;
}
int isprime(int n)
{
int i;
for(i = 2;i < n;i++)
{
if((n%i)==0)
{
break;
}
}
if(i == n)
{
return 1;
}
else
{
return 0;
}
}
10、对二维数组排序，数组中元素最小的元素排在第一行 ，依次排完（后期应该会单独出一期排序算法 ，现在排序写的是有点复杂）
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#define N 6
int main()
{
void Output(int a[][N]);
void sort(int *p[N]);
int min(int *p);
srand((unsigned)time(NULL));
int i,j;
int a[N][N];
int *p[N];
int *q;
for(i = 0;i < N;i++)
{
for(j = 0;j < N;j++)
{
a[i][j] = rand()%100;
}
}
Output(a);
for(i = 0;i < N;i++)
{
p[i] = a[i];
}
sort(p);
for(i = 0;i < N;i++)
{
q = p[i];
for(j = 0;j < N;j++)
{
printf("%4d",*q);
q++;
}
printf("\n");
}
printf("\n");
return 0;
}
void Output(int a[][N])
{
int i;
int j;
for(i = 0;i < N;i++)
{
for(j = 0;j < N;j++)
{
printf("%4d",a[i][j]);
}
printf("\n");
}
printf("\n");
}
int min(int *p)
{
int i = 0;
int min;
min = *p;
for(i = 0;i < N;i++)
{
if(min > *p)
{
min = *p;
}
p++;
}
return min;
}
void sort(int *p[N])
{
int i,j,k;
int *q;
for(i = 0; i < N-1;i++)
{
for(j = i;j < N;j++)
{
if(min(p[i])>min(p[j]))
{
q = p[i];
p[i] = p[j];
p[j] = q;
}
}
}
}
标签：语言,int,void,char,++,printf,习题,include,529 From： https://www.cnblogs.com/gunancheng/p/17444638.html