数字频率程序
请帮助我使用此代码以获得清晰的输出。我是初学者
我已经使用数组制作了程序。我不知道这是否正确。用我自己的逻辑制作
int count(int a)
{
int c;
while(a>=1)
{
c++;
a=a/10;
}
return c;
}
int main()
{
//program to find frquency of the number
int a,n,d;
int b[100];
int e[100];
scanf("%d",&a);
n=count(a);
for(int i=n;a>0;i--)
{
b[i]=a%10;
a=a/10;
}
for(int i=1;i<=n;i++)
{
d=b[i];
e[d]++;//most probably this part error occurs
printf("%d\n",d); //used this this to confirm that i have correctly stored value in d.
}
for(int i=1;i<=n;i++)
{
printf("%d ",e[i]);
}
return 0;
}
答案 0 :(得分:0)
int c;
行应为int c = 0;
int e[100];
行应为int e[100] = {0};
以下code
可以工作:
#include <stdio.h>
int count(int a) {
int c = 0;
while (a >= 1) {
c++;
a = a / 10;
}
return c;
}
int main() {
// program to find frquency of the number
int a, n, d;
int b[100];
int e[100] = {0};
scanf("%d", &a);
n = count(a);
for (int i = n; a > 0; i--) {
b[i] = a % 10;
a = a / 10;
}
for (int i = 1; i <= n; i++) {
d = b[i];
e[d]++; // most probably this part error occurs
printf("%d\n", d); // used this this to confirm that i have correctly
// stored value in d.
}
for (int i = 1; i <= n; i++) {
printf("%d ", e[i]);
}
return 0;
}
此外,您可以使用snprintf
:
#include <stdio.h>
int main() {
int a;
int max = -1;
char buf[100];
int count[10] = {0};
scanf("%d", &a);
snprintf(buf, sizeof(buf), "%d", a);
for (int i = 0; buf[i] != '\0'; ++i) {
int temp = buf[i] - '0';
++count[temp];
if (temp > max)
max = temp;
}
for (int i = 0; i <= max; ++i)
printf("%d ", count[i]);
return 0;
}