The task is simple: given any positive integer , you are supposed to count the total number of 1’s in the decimal form of the integers from 1 to . For example, given being 12, there are five 1’s in 1, 10, 11, and 12.
Input Specification:
Each input file contains one test case which gives the positive .
Output Specification:
For each test case, print the number of 1’s in one line.
Sample Input:
12
Sample Output:
5
题意
计算0-n中出现的"1"的次数。
思路
对每一位分别计算为"1"的次数,left为高位组成的数,right为低位组成的数, base为这位的基(10^(n - i))
当这位为0时,left * a;(高位为0 ~ left - 1时,这位可能出现1)
当这位为1时,left * a + right + 1 (高位为0 ~ left - 1时,这位可能出现1;当高位为left,且低位为0 ~ right时,这位可以是1)
当这位大于1时,(left + 1) * a (高位为0 ~ left时,这位可能出现1)
代码
#include <cstdio>
int main() {
int n, a = 1, ans = 0;
int left, now, right;
scanf("%d", &n);
while (n / a != 0) {
left = n / (a * 10);
now = n / a % 10;
right = n % a;
if (now == 0)
ans += left * a;
else if (now == 1)
ans += left * a + right + 1;
else
ans += (left + 1) * a;
a *= 10;
}
printf("%d\n", ans);
}