C语言 求大神改成正确的 题目是:输入两个整数num1和num2,计算并输出它们的和、差、积、商和余数#include int main(void){ int num1,num2; int x,y,z,k,j;print("Enter num1=",num1);scanf("%d",&num1);print("Enter num2=",nu
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![C语言 求大神改成正确的 题目是:输入两个整数num1和num2,计算并输出它们的和、差、积、商和余数#include int main(void){ int num1,num2; int x,y,z,k,j;print(](/uploads/image/z/5208470-62-0.jpg?t=C%E8%AF%AD%E8%A8%80+%E6%B1%82%E5%A4%A7%E7%A5%9E%E6%94%B9%E6%88%90%E6%AD%A3%E7%A1%AE%E7%9A%84+%E9%A2%98%E7%9B%AE%E6%98%AF%EF%BC%9A%E8%BE%93%E5%85%A5%E4%B8%A4%E4%B8%AA%E6%95%B4%E6%95%B0num1%E5%92%8Cnum2%2C%E8%AE%A1%E7%AE%97%E5%B9%B6%E8%BE%93%E5%87%BA%E5%AE%83%E4%BB%AC%E7%9A%84%E5%92%8C%E3%80%81%E5%B7%AE%E3%80%81%E7%A7%AF%E3%80%81%E5%95%86%E5%92%8C%E4%BD%99%E6%95%B0%23include+int+main%28void%29%7B+int+num1%2Cnum2%3B++int+x%2Cy%2Cz%2Ck%2Cj%3Bprint%28%22Enter+num1%3D%22%2Cnum1%29%3Bscanf%28%22%25d%22%2C%26num1%29%3Bprint%28%22Enter+num2%3D%22%2Cnu)
C语言 求大神改成正确的 题目是:输入两个整数num1和num2,计算并输出它们的和、差、积、商和余数#include int main(void){ int num1,num2; int x,y,z,k,j;print("Enter num1=",num1);scanf("%d",&num1);print("Enter num2=",nu
C语言 求大神改成正确的 题目是:输入两个整数num1和num2,计算并输出它们的和、差、积、商和余数
#include
int main(void)
{ int num1,num2;
int x,y,z,k,j;
print("Enter num1=",num1);
scanf("%d",&num1);
print("Enter num2=",num2);
scanf("%d",&num2);
x=num1+num2;
y=num1-num2;
z=num1*num2;
k=num1/num2;
j=num1%num2;
print("num1+num2=x",x);
print("num1-num2=y",y);
print("num1*num2=z",z);
print("num1/num2=k",k);
print("num1%num2=j",j);
return 0;
}
我上C语言就4节课 新手
我把下面改成printf("num1+num2=%d",x);
printf("num1-num2=%d",y);
printf("num1*num2=%d",z);
printf("num1/num2=%d",k);
printf("num1%num2=%d",j);
提醒成功编译,但有警告信息.
xfc2_6.c: In function ‘main’:
xfc2_6.c:5: warning: too many arguments for format
xfc2_6.c:7: warning: too many arguments for format
xfc2_6.c:18: warning: format ‘%n’ expects type ‘int *’, but argument 2 has type ‘int’
xfc2_6.c:18: warning: too few arguments for format
有什么要改的吗? 谢谢
额 最后格式还要这样的
Enter num1:5
Enter num2:3
5+3=8
5-3=2
5*3=15
5/3=1.67
5%3=2
C语言 求大神改成正确的 题目是:输入两个整数num1和num2,计算并输出它们的和、差、积、商和余数#include int main(void){ int num1,num2; int x,y,z,k,j;print("Enter num1=",num1);scanf("%d",&num1);print("Enter num2=",nu
print改成printf就可以了
标准答案 拿去吧 除法进行的Int型的除法 自己懂得
#include
int main(void)
{ int num1,num2;
int x,y,z,k,j;
printf("Enter num1=");
scanf("%d",&num1);
printf("Enter num2=");
scanf("%d",&num2);
x=num1+num2;
y=num1-num2;
z=num1*num2;
k=num1/num2;
j=num1%num2;
printf("num1+num2=%d\n",x);
printf("num1-num2=%d\n",y);
printf("num1*num2=%d\n",z);
printf("num1/num2=%d\n",k);
printf("num1%%num2=%d\n",j);
return 0;
}