1、倒着打印
#include<stdio.h>
void main()
{
short score[]={1,2,3,4,5};
int i=0;
for (i=4;i>=0;i-- )
{
printf("%d\n",score[i]);
}
}
2、判断某个值是否存在
#include<stdio.h>
void main()
{
short score[]={1,2,3,4,5};
int i=0,j=0;
scanf("%d",&j);
for ( i=0;i<5;i++ )
{
if (j == score[i] )
{
printf("yes");
break;
}
if (i == 4 )
{
printf("no");
}
}
3、此数组元素的和
#include<stdio.h>
void main()
{
short score[]={1,2,3,4,5};
int i=0,sum=0;
for(i=0;i<5;i++ )
{
sum += score[i];
}
printf("%d\n",sum);
}
4、输出最大值的下标和元素值
#include<stdio.h>
void main()
{
short score[]={1,2.3,4,5};
int i=0,j=0;
int listen=score[0];
for(i=1;i<5;i++)
{
if( listen < score[i] )
{
listen=score[i];
j=i;
}
}
printf("%d%d\n",listen,j);
}
5、char buf[6]={‘l’,‘v’,‘q’,‘i’,‘n’,‘g’}输出
#include<stdio.h>
void main()
{
int i=0;
for (i=0;i<6;i++)
{
printf("%c\n",buf[i]);
}
}
6、求buf的字符数量。
#include<stdio.h>
void main()
{
char name[]={1,2,3,4,5,6,7};
int i=0;
int j=0;
while(name[i] != 0)
{
i++
}
printf("%d\n",i);
printf("%d\n",sizeof(name));
printf("%d\n",strlen(name));
}
7、char buf[]=“100a” int num;
#include<stdio.h>
void main()
{
char buf[]="100a";
int i=0,j=0,num=0;
while( buf[i] != '\0' )
{
if ( buf[i] >='0' && buf[i] <='9' )
{
printf("%c",buf[i]);
}
i++;
}
}
8、char buf[]=“fdsaKFDU*&^*9889f763JKJFDK”
统计:大写,小写,数字,其他字符数量
#include<stdio.h>
void main()
{
int i=0,a=0,b=0,c=0,d=0,f=0,m=0;
while( buf[i] != '\0' )
{
if ( buf[i] >= 'A' && buf[i] <= 'Z')
{
a=a+1;
}
else if ( buf[i] >= 'a' && buf[i] <= 'Z' )
{
m=buf[i] - 32;
printf("%c\t",m);
b=b+1;
}
else if ( buf[i] == 32 )
{
c=c+1;
}
else if ( buf[i] >='0' && buf[i] <='9' )
{
d=d+1;
}
else
f=f+1;
i++;
}
printf("%d\n",a);
printf("%d\n",b);
printf("%d\n",c);
printf("%d\n",d);
printf("%d\n",f);
}
9、选择排序
#include<stdio.h>
void main()
{
short score[]={9,4,6,3,7,2,5,1};
int i=0,j=0;// i未排序 j用于取最值
int listen=0;//监哨值法
int temp=0;
for ( i=0;i<7-1;i++ )
{
listen=i;
for( j=i+1;j<7;j++ )
{
if( score[listen] < score[j] )
listen=j;
}
//两数对换
if( i != listen )
{
temp = score[i];
score[i]=score[listen];
score[listen]=temp;
}
}
for( i=0;i<7;i++ )
printf("%d\n",score[i]);
}