Question

用C语言编写计算100至1000之间有多少个数其各位数字之和是5的程序是什么呀

这里该填什么？main()
{ int i,s,k,count=0;
for(i=100;i<=1000;i++)
{ s=0;k=i;
while(___1__)
{ s=s+k%10;
k=___2___;
}
if(s!=5) ___3___;
else count++;
}
printf("%d",count);
}

Answer 1

都过了很久了^_^

测试程序如下：

#include<iostream.h> 

        int main()   

{    int i, s, k,count=0;  

for(i=100;i<1000;i++)  

{  s=0; k=i;              

while(k>0)     

{ s=s+k%10;           

  k=k/10;}       

if(s!=5)

continue; 

    else 

count++;         

}            

cout<<count<<endl;        

}  

测试通过

Answer 2

#include
<stdio.h>
int
main(void)
{
int
i,s,k,count=0;
for(i=100;i<=1000;i++)
{
s=0;k=i;
while(k)
{
s=s+k%10;
k=k/10;
}
if(s!=5)
continue;
else
count++;
}
printf("%d",count);
return
0;
}
#include
<stdio.h>
int
getSum(int
num)
{
char
buf[10];
int
sum
=
0;
int
i
=
0;
sprintf(buf,"%d",num);
while(buf[i])
sum
+=
buf[i++]
-
'0';
return
sum;
}
int
main(void)
{
int
i;
for(i
=
100;
i
<=
1000;
i++)
if(getSum(i)
==
5)
printf("%d\t",i);
return
0;
}

Answer 3

1.k>0
2.k/10
3.break

Answer 4

#include <stdio.h>
int main(void)
{
int i,s,k,count=0;
for(i=100;i<=1000;i++)
{ s=0;k=i;
while(k)
{ s=s+k%10;
k=k/10;
}
if(s!=5) continue;
else count++;
}
printf("%d",count);

return 0;
}

#include <stdio.h>
int getSum(int num)
{
char buf[10];
int sum = 0;
int i = 0;
sprintf(buf,"%d",num);
while(buf[i])
sum += buf[i++] - '0';
return sum;
}
int main(void)
{
int i;
for(i = 100; i <= 1000; i++)
if(getSum(i) == 5)
printf("%d\t",i);
return 0;
}