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
Solution
数学
设\(N=d_nd_{n-1}d_{n-2}\cdots d_i \cdots d_2d_1\)，对于每一个数位\(d_i\)，计算当前数位可能出现1的次数并累加起来即可获得答案。
假设当前位为\(cur \rightarrow d_i\)，记其左边的数\(left=d_nd_{n-1}d_{n-2}\cdots d_{i+1}\)，右边的数为\(right=d_{i-1} \cdots d_3d_2d_1\)。使用一个变量\(a\)来记录当前位是个位，十位还是百位类推。根据数位的变化规律，有如下结论：
- \(cur==0\)，\(ans+=left*a\)。当高位的数值从\(0\)变换到\(left-1\)时，在\(cur\)位处会出现\(left\)次1，因为对于一个数\(d_nd_{n-1}d_{n-2}\cdots d_i XXXX\)来说，前缀一旦确定，那么\(d_i\)只会在当前前缀的数字下出现一次。但是由于\(cur==0\)，所以在以\(left\)为前缀的数中，\(right\)从\(0\)到\(999\cdots\)，因此对于一个确定的\(left\)，\(d_i\)出现1的次数有\(a\)次，因此一共有\(left*a\)次1出现。
- \(cur>1\),\(ans+=(left+1)*a\)。由于当前位的值大于1，因此高位的数值可以从\(0\)变换到\(left\)，在\(cur\)位处会出现\(left+1\)次\(1\).
- \(cur==1\)，\(ans+=left*a+right+1\)，\(left*a\)的来源与\(cur==0\)的情况一致。由于\(cur==1\)，因此当左侧前缀\(left\)取到最大时，右侧的后缀的范围不是从0~999...，而是从\(0\)到\(right(d_{i-1} \cdots d_3d_2d_1)\)，因此这里还有\(right+1\)个1.
#include<bits/stdc++.h>
long long ans=0;
int main(){
int N;
std::cin>>N;
int left=0,right=0,cur=1,a=1;
int ans=0;
while(N/a){
left=N/(a*10);
right=N%a;
cur=N/a%10;
if(cur==0) ans+=left*a;
else if(cur==1) ans+=left*a+right+1;
else if(cur>1) ans+=(left+1)*a;
a*=10;
}
std::cout<<ans;
return 0;
}