我必须计算某个数字在某个范围内的每个数字中重复的次数。例如，在 0 到 20 之间的数字中，仅出现一次 1 重复两次 (11)。我最初通过将 int 转换为 str 并迭代它来完成此操作，但我希望能够以算术方式解决此问题。有什么想法吗？
您可以将您的数字多次除以 10：
int number = 72;
int rest;
while(number)
{
rest = number % 10;
printf("%d\n", rest);
number /= 10;
}
这里 rest 包含“2”，然后包含“7”
这是您可以使用的通用解决方案，您的问题没有包含太多信息，所以我假设您想要计算每个数字中每个数字的重复次数。
所以我所做的就像散列，其中每个数字中的数字永远不会交叉值
909arrarr例如，数字
554arr[5]++;arr[4]++;最后我迭代哈希数组，打印每个数字中每个数字出现的次数。
这是代码：
#include <stdio.h>
#include <math.h>
int main()
{
int arr[6] = {5555, 12112, 14, -3223, 44, 10001};
int tempArr[10] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
for (int i = 0; i < 6; i++) {
int temp1 = arr[i];
// get the number of occurrences of each digit in each number
do{
tempArr[(abs(temp1) % 10)]++;
temp1 /= 10;
}while(temp1 != 0);
// loop over the array to know how many times each digit occurred
printf("the number of occurrences in number called %d\n", arr[i]);
for (int j = 0; j < 10; j++) {
if(tempArr[j] > 1)
printf("number %d occurred %d times\n", j, tempArr[j]);
// resetting that position of the array
tempArr[j] = 0;
}
}
return 0;
}
这是输出：
the number of occurrences in number called 5555
number 5 occurred 4 times
the number of occurrences in number called 12112
number 1 occurred 3 times
number 2 occurred 2 times
the number of occurrences in number called 14
the number of occurrences in number called -3223
number 2 occurred 2 times
number 3 occurred 2 times
the number of occurrences in number called 44
number 4 occurred 2 times
the number of occurrences in number called 10001
number 0 occurred 3 times
number 1 occurred 2 times
某个数字在某个范围内的每个数字中重复了多少次。
伪代码*1
获取范围：
imin, imaxintimin <= imax获取数字：
digit从
miminimax....打印
m....设置
digit_count = 0...重复
.......
ldmabs(m%10)....... 如果
ld == digitdigit_count........ 将
m....直到
m == 0....打印
digit_count完成
*1 由于OP没有提供代码，似乎最好不要提供代码答案。
重复 @Abdo Salm 提供的答案只是为了演示一种稍微另类的方法。一切都归功于阿卜杜。
#include <stdio.h>
#include <string.h>
#include <limits.h>
void process( int n, int cnts[] ) {
// count occurrences of each digit (right to left)
while( n )
cnts[ abs(n % 10) ]++, n /= 10;
}
void report( int cnts[], int thresh ) {
char *sep = "";
for( int j = 0; j < 10; j++ )
if( cnts[ j ] > thresh )
printf( "%s%dx%d", sep, cnts[ j ], j ), sep = ", ";
if( !*sep )
printf( "no digit occurred multiple times" );
putchar( '\n' );
}
int main( void ) {
int ranges[][2] = {
{ 0, 10, },
{ 0, 22, },
{ 110, 133, },
{ 447, 448, },
};
// four trail ranges above
for( int r = 0; r < sizeof ranges/sizeof ranges[0]; r++ ) {
int metacnts[ 10 ] = {0};
// examine each number in the range (inclusive of start & end)
for( int i = ranges[r][0]; i <= ranges[r][1]; i++ ) {
int cnts[ 10 ] = {0};
process( i, cnts );
// tabulate only 'digits' occurring more than once
for( int j = 0; j < 10; j++ )
if( cnts[ j ] > 1 )
metacnts[ j ]++;
}
// report only digits appearing multiple times in numbers between min & max
printf( "Range %3d-%3d (incl) ", ranges[r][0], ranges[r][1] );
report( metacnts, 0 );
}
return 0;
}
输出：
Range 0- 10 (incl) no digit occurred multiple times
Range 0- 22 (incl) 1x1, 1x2
Range 110-133 (incl) 12x1, 1x2, 1x3
Range 447-448 (incl) 2x4