c语言求解答
c\u8bed\u8a00\u6c42\u89e3\u7b54/*
3/6
3 + 6 = 0.5
2*3
2 + 3 = 6
2/0
\u9664\u6570\u4e0d\u80fd\u4e3a\u96f6\u3002
135*2
135 + 2 = 270
q
Press any key to continue
*/
#include int main() {int a,b;char op;while(scanf("%d%c%d",&a,&op,&b) == 3) {switch(op) {case '+' : printf("%d + %d = %d\n",a,b,a + b); break;case '-' : printf("%d + %d = %d\n",a,b,a - b); break;case '*' : printf("%d + %d = %d\n",a,b,a * b); break;case '/' : if(b) printf("%d + %d = %g\n",a,b,1.0 * a / b);else printf("\u9664\u6570\u4e0d\u80fd\u4e3a\u96f6\u3002\n"); break;default : printf("\u4e0d\u80fd\u8bc6\u522b\u7684\u8fd0\u7b97\u7b26:%c\n",op);}}return 0;}
main()
{
int i,m,n,j,k;
int a[10];
m=0;n=0;j=0;k=0;
for(i=0;i<10;i++)
{
scanf("%d",&a[i]);
if(a[i]%2==0)
{
m+=a[i];
j+=1;
}
else
{
n+=a[i];
k+=1;
}
}
printf("\u5947\u6570\u5e73\u5747\u503c=%d",m/j);
printf("\u5076\u6570\u5e73\u5747\u503c=%d",n/k);
}
第一个答案:
#include<stdio.h>
void delchar(char * str, char c)
{
int i = 0;
if(str[0] =='\0')
return ;
if(str[0] == c)
{
for(i=0;str[i] !='\0';i++)
{
str[i] = str[i+1];
}
}
delchar(str+1,c);
}
int main()
{
char str[20];
char c;
gets(str);
c = getchar();
delchar(str,c);
puts(str);
return 0;
}
第三题答案
#include<stdio.h>
char * str_bin(char * str1,char * str2)
{
char result[200];
int i=0;
int j=0;
int k=0;
while(str1[i]!='\0' && str2[j] != '\0')
{
if(str1[i]<str2[j])
{
result[k++] = str1[i++];
}
else
{
result[k++] = str2[j++];
}
}
while(str1[i]!='\0')
{
result[k++] = str1[i++];
}
while(str2[j]!='\0')
{
result[k++] = str2[j++];
}
result[k] = '\0';
return result;
}
int main()
{
char str1[100];
char str2[100];
char * result;
gets(str1);
gets(str2);
result = str_bin(str1,str2);
puts(result);
return 0;
}
第四个答案
#include<stdio.h>
int judge(char * str)
{
int i = 0;
char c1 = '0';
char c2 = '0';
if(str[0] == '\0')
return -1;
if(str[1] == '\0')
{
c1 = str[0];
}
for(i = 1;str[i] != '\0';i++)
{
c2 = str[i-1];
c1 = str[i];
}
if(c1 =='s' || c1 == 'x' || c1 == 'o')
return 2;
if(c1 == 'h' &&(c2 == 'c' || c2 == 's'))
return 2;
if(c1 == 'y')
return 3;
return 1;
}
void toPl(int state,char *str)
{
int i=0;
switch(state)
{
case 1:
printf("%s",str);
puts("s");
break;
case 2:
printf("%s",str);
puts("es");
break;
case 3:
for(i = 1; str[i] != '\0';i++)
{
putchar(str[i-1]);
}
puts("ies");
}
}
int main()
{
char str[100];
gets(str);
toPl(judge(str),str);
return 0;
}
所有的答案都没写注释,不过代码都没有问题,我简单的测试了一下,如果有问题的话加QQ1336400462
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