用C语言写出输入一个三位的整数,计算这个数各位数值之和. 谁知道 要准确 高人来 C语言：输入一个三位的整数，计算其每位数字的累加之和。
\u7528C\u8bed\u8a00\u5199\u51fa\u8f93\u5165\u4e00\u4e2a\u4e09\u4f4d\u7684\u6574\u6570,\u8ba1\u7b97\u8fd9\u4e2a\u6570\u5404\u4f4d\u6570\u503c\u4e4b\u548c. \u8c01\u77e5\u9053#include
int main()
{
printf("\u8f93\u5165\u4e00\u4e2a\u4e09\u4f4d\u7684\u6574\u6570");
int num;
scanf("%d", &num);
int sum = 0;
for (int i=0;i<3;i++)
{
sum += num%10;
num/=10;
}
printf("\u8fd9\u4e2a\u6570\u5404\u4f4d\u6570\u503c\u4e4b\u548c\u4e3a%d", sum);
while (1)
;
return 0;
}
#include
#include
int main()
{
int a,x,y,s,sum;
printf("\u8f93\u5165\u4e00\u4e2a\u4e09\u4f4d\u6574\u6570\uff1a");
scanf("%d",&a);
x=a/100; //a\u9664\u4ee5100\u53d6\u6574\uff0c\u5c31\u662f\u53d6\u767e\u4f4d\u6570
y=a/10%10; //a\u9664\u4ee510\u53d6\u6574\u518d\u9664\u4ee510\u53d6\u4f59\uff0c\u5c31\u662f\u53d6\u5341\u4f4d\u6570
s=a%10; //a\u9664\u4ee510\u53d6\u4f59\uff0c\u5c31\u662f\u4e2a\u4f4d\u6570
sum=x+y+s; //\u4e09\u4e2a\u6570\u4f4d\u76f8\u52a0\u4e4b\u548c
printf("\u6bcf\u4f4d\u6570\u5b57\u7d2f\u52a0\u4e4b\u548c\u4e3asum=%d",sum);
return 0;
}
#include<stdio.h>
void main()
{
int num;
int ge ;
int shi ;
int bai ;
int s=0;
printf("请输入一个三位数\n");
scanf("%d",&num);
if(num>=100&&num<=999)
{
ge=num%10;
shi = num/10%10;
bai = num/100;
s= ge+shi+bai;
printf("三位数各位之和为%d\n",s);
}
else
printf("输入错误，请重新输入！") ;
}
void main()
{
int a = 123;
int ge =a%10;
int shi = a/10%10;
int bai = a/10;
int s= ge+shi+bai;
}
其它部分自己补
int a,sum=0;
scanf("%d",&a);
while(a)
{sum=sum+a%10;
s/=10;}
printf("%d",sum);
有解了·····
最笨的方法：
void main()
{
int sum=0,a;
printf("please input number:\n");
scanf("%d",&a);
sum=sum+a/100;
sum=sum+a/10%10;
sum=sum+a%10;
printf("sum=%d",sum);
getch();
}
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