PAT甲级1049(数1)

1049 Counting Ones (30 point(s))

  The task is simple: given any positive integer N, you are supposed to count the total number of 1’s in the decimal form of the integers from 1 to N. For example, given N 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 N (≤2^​30).

Output Specification:
  For each test case, print the number of 1’s in one line.

Sample Input:

12

Sample Output:

5

题目大意

  给定一个数字N，要求你计算从1-N的所有数字中出现了多少个1

解题思路

  这个题我没做出来，看的算法笔记的思路
从低位到高位枚举每一位数字，同时得到当前数字左边数字的值left和右边数字的值right。如果当前数字为0，则当前位出现1的次数为左边0~left-1=left乘a；若当前数字为1，则当前位出现1的次数为left*a+right+1；若当前数字大于1，则当前位出现1的次数为(left+1)*a，详见代码

代码

#define _CRT_SECURE_NO_WARNINGS
#include 

//此题采用暴力法很显然会超时，直接累加每个位置上出现1的次数即可得最终结果

int main()
{
	int N;
	int ret = scanf("%d", &N);
	int left, now, right, res = 0, a = 1;//left代表当前数字左边的值，right代表右边的值，now代表当前数字
	while (N / a) {		
		left = N / (a * 10);
		now = N / a % 10;
		right = N % a;
		if (now == 0) {//如果当前数字等于0，则当其左边0到left-1，及其右边随便取值都可使当前位置为1
			res += left * a;
		}
		else if (now == 1) {//如果当前数字为1，左边0到left-1及右边随便取值加上当前位为1时，右边可取0到right
			res += left * a + right + 1;
		}
		else res += (left + 1) * a;//如果当前数字大于1，左边0到left及右边随便取值
		a *= 10;
	}
	printf("%d", res);
	return 0;
}