[英]Sum of digits of some number
例如,如果我输入 12345 那么 sum = 1+2+3+4+5
int i ;
int sum;
int individual;
int n;
printf("enter the number: ");
scanf("%d",&n);
for (i=0;i<5;++i){
**indvidual=+n[i];**
sum=+invidual;
}
printf("%d",sum)
return 0;
您可以除以 10 并查看余数以获得每个数字。
此外,您应该使用+=
而不是=+
将内容添加到变量并在使用它们的值之前初始化变量。
int i ;
int sum = 0; /* initialize */
int individual;
int n;
printf("enter the number: ");
scanf("%d",&n);
for (i=0;i<5;++i){
indvidual=+(n%10); /* obtain a digit (+ is not required, but left by respect) */
n/=10; /* eliminate last dight and proceed to next digit */
sum+=invidual; /* use += instead of =+ */
}
printf("%d",sum); /* also add semicolon here */
return 0;
#include <stdlib.h>
#include <stdio.h>
int main()
{
int i = 12345;
int sum;
i = abs(i);
sum = i?i%9?i%9:9:0;
printf ("Sum of digits is %d\n", sum);
}
因为1 + 2 + 3 + 4 + 5 = 15
然后1 + 5 = 6
(或者是预期的 output 15
,只需通过数字一次?)
#include <stdlib.h> #include <stdio.h> int main() { int i = 12345; int sum=0; i = abs(i); while(i) { sum += i%10; i /= 10; } printf ("Sum of digits is %d\n", sum); return 0; }
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