PAT-A1049 Counting Ones 题目内容及题解

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

题目给定任意正整数N，要求输出从1到N的整数的十进制数中的“1”的总数。

解题思路

1. 从个位开始分别计算每一位上1的数量，并将其累加；
2. 输出结果并返回零值。

代码

#include
int main(){
    int N,a=1,ans=0;
    int left,now,right;
    scanf("%d",&N);
    while(N/a){
        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);
    return 0;
}

运行结果