This question requires a simple statistical function integer number of specified numbers.
Function interface definition:
int CountDigit( int number, int digit );
Wherein number
is an integer of not more than long integer, digit
is an integer within [0, 9] interval. Function CountDigit
should return number
the digit
number of occurrences.
Referee test program Example:
#include <stdio.h>
int CountDigit( int number, int digit );
int main()
{
int number, digit;
scanf("%d %d", &number, &digit);
printf("Number of digit %d in %d: %d\n", digit, number, CountDigit(number, digit));
return 0;
}
/* 你的代码将被嵌在这里 */
Sample input:
-21252 2
Sample output:
Number of digit 2 in -21252: 3
answer:
int CountDigit( int number, int digit ){
int a[10]={0};
if(number<0) number= -1*number;
while(number>10) {
a[number-(number/10)*10]++;
number=number/10;
}
a[number]++;
return a[digit];
}