1049 Counting Ones （30 分）

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：给出一个数字n，求1~n的所有数字里面出现1的个数

题解：第一种方法是，扫每一位，然后把每一位按是不是1来考虑，很好很实用

#include <iostream>
using namespace std;
int main() {
    int n, left = 0, right = 0, a = 1, now = 1, ans = 0;
    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 = a * 10;
    }
    printf("%d", ans);
    return 0;
}

 第二种颇有些数位dp的意思，找转移方程，反正我有些佩服，不过但凡是dp都会，有空间上的消耗

#include<iostream>
#include<vector>
using namespace std;
int main(){
    long long int n,sum=0;
    cin>>n;
    int a[n+1]={0};
   //vector<int> a(n+1,0);
    for(int i=1;i<=n;i++){
        if(i%10==0)
            a[i]=a[i/10];
        else if(i%10==1)
            a[i]=a[i-1]+1;
        else if(i%10==2)
            a[i]=a[i-1]-1;
        else 
            a[i]=a[i-1];
        sum+=a[i];
    }
    cout<<sum<<endl;
    return 0;
}