PAT 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 (<=2^30^).

Output Specification:

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

Sample Input:

12

Sample Output:

5

#include<iostream>
#include<vector>
using namespace std;
int main(){
  int n;
  while(1){
  cin>>n;
  if(n==-1) break;
  vector<int> v;
  v.push_back(0);
  for(int i=1; i<=n; i++){
    int cnt=0, temp=i;
    while(temp){
        if(temp%10==1) cnt++;
        temp/=10;
    }
    v.push_back(v[i-1]+cnt);
  }
  cout<<v[n]<<" "<<f(n)<<endl;
  }
  return 0;
}

 

观察可以发现每一位1出现的次数和当前位左边的数有关 也和右边的数有关；
计算1的个数还与当前为的数有关
当前位为0的时候
当前位为1
当前位为12..9

#include<iostream>
#include<vector>
#include<cmath>
using namespace std;
 
int main(){
  int n;
  cin>>n;
  int cnt=0, cpy=n;
  vector<int> v;
  while(cpy){//求n的每一位数
    v.push_back(cpy%10);
    cpy /= 10;
  }
  cnt = n/10;
  if(n%10>=1) cnt++;
  int idx=2, coe=10;
  while(idx<=v.size()){
    cnt += (n/(coe*10))*coe;
    if(v[idx-1]>1)  cnt += coe;
    else if(v[idx-1]==1) cnt += (n%coe+1);
    coe *= 10;
    idx++;
  }
  cout<<cnt;
  return 0;
}