数字频率的程序
请帮助我编写此代码以获得清晰的输出。我是初学者
我已经使用数组编写了该程序。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;
}
最佳答案
- 行
int c;
应为int c = 0;
- 行
int e[100];
应为int e[100] = {0};
以下代码
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,您还可以使用 snprintf
来实现:
#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;
}
关于c - 为什么这个程序的输出是错误的?.数字的频率,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54007736/