1049. Counting Ones (30)-PAT甲级真题

1049. Counting Ones (30)
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 (<=230).

Output Specification:

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

Sample Input:
12
Sample Output:
5

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　　分析：这道题就是求１到ｎ之间有的数字里面有多少次出现１，那我们就一位一位判断，left表示当前为左边的所有，right表示当前位右边的所有，每次去判断now位是否是1的时候，和左边右边没有关系，他们能随便取值，不会出现重复，那么当前的这一位的情况ｎｏｗ：
　　1.now = 0 : 那么 ans += left * a; 因为now=0说明now位只有在left从0~left-1的时候会产生1，所以会产生left次，再看右边，右边可以去0~a的任意值，所以有a个
　　2.now = 1 : ans += left * a + right + 1;now = 1的时候就要比上一步多加一个当now为1的时候右边出现0~right个数导致的now为1的次数
　　3.now >= 2 : ans += (left + 1) * a;now大于等于2就左边0~left的时候会在now位置产生1，所以会产生left次，同样右边也是取0~a的任意值
　　最后想说的是这道题求的是1出现的次数，不是包含1数的个数，所以11这种算2.。。。。。好坑

#include <cstdio>using namespace std;int main (){    int n , left = 0 , right = 0 , a = 1 , now = 1 , result = 0;    scanf_s ("%d" , &n);    while (n / a)    {        left = n / (a * 10) , now = n / a % 10 , right = n % a;        if (now == 0)            result += left * a;        else if (now == 1)            result += left * a + right + 1;        else             result += (left + 1) * a;        a = a * 10;    }    printf ("%d" , result);    return 0;}