题目描述
A Benedict monk No. 16 writes down the decimal representations of all natural numbers between and including
m and n , m≤ n
. How many 0's will he write down?
Input consists of a sequence of lines. Each line contains two unsigned 32-bit integers m and n, m ≤ n. The last line of input has the value of m negative and this line should not be processed.
For each line of input print one line of output with one integer number giving the number of 0's written down by the monk.
输入
暂无
输出
暂无
样例输入
10 11
100 200
0 500
1234567890 2345678901
0 4294967295
-1 -1
样例输出
1
22
92
987654304
3825876150
参考代码
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <assert.h>
unsigned countl(char *s)
{
unsigned r;
if (strlen(s) == 1) return 1;
if (s[0] == '0')
{
sscanf(s+1,"%lu",&r);
return 1 + r + countl(s+1);
}
r = pow(10,strlen(s+1));
r += (s[0]-'0') * strlen(s+1) * pow(10,strlen(s+1)-1);
r += countl(s+1);
return r;
}
unsigned count(char *s)
{
unsigned r = 1;
char tmp[20];
if (strlen(s) == 1) return 1;
for (r=0;r<strlen(s)-1;r++) tmp[r] = '9';
tmp[r] = 0;
r = count(tmp);
r += (s[0]-'1') * strlen(s+1) * pow(10,strlen(s+1)-1);
r += countl(s+1);
return r;
}
char x[100], y[100];
unsigned u;
int main()
{
while (2 == scanf("%s%s",x,y) && x[0] != '-')
{
if (x[0] == '0') printf("%lun",count(y));
else {
sscanf(x,"%lu",&u);
sprintf(x,"%lu",u-1);
printf("%lun",count(y)-count(x));
}
}
return 0;
}
解析
暂无
