本文已参与「新人创作礼」活动，一起开启掘金创作之路
题目
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
题解
这道题的思考过程不难，只要看过就能理解这个过程，从而代码实现出来。题目本身很新颖，但背后也没有特别深刻的思想体系。
这道题的核心是：针对个、十、百、千每一位出现1的次数，对数字1进行统计。然后只要在纸上写一下每一位1的个数是怎么统计的，就知道如何写统计的代码了。
具体的逻辑是要看这一位的值，如果小于1或大于1，则这一位1的个数是整的，由高位；如果是1则要额外考虑高位相等同位是1的数的个数，这个个数由低位确定。
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <cstdio>
#include <bitset>
using namespace std;
const int MAX = 100005;
const int INF = 0x3f3f3f3f;
int count_num(int n_in, int num){
int n = n_in;
int result = 0;
int ten = 1;
while(n > 0){
int tail = n % 10;
n /= 10;
if(tail < num){
result += n * ten;
}
else if(tail == num){
result += n * ten + (n_in - n * ten * 10 - ten + 1);
}
else{
result += (n + 1) * ten;
}
ten *= 10;
}
return result;
}
int count1(int n){
return count_num(n, 1);
}
int main(){
int n;
cin >> n;
cout << count1(n);
return 0;
}
