PAT 1049. Counting Ones (30)（数位dp（记忆化搜索））

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
题意很好理解，求1到n这n个数其中1总共出现了多少次。
还是比较基础的数位dp，dp数组建议使用结构体，保存两个量，1出现了多少次和总共有多少个数。至于为什么呢，也是很好理解的，如果当前位不是1，那么总共出现的1的个数和当前位减一出现的1的个数相等，如果当前位是1，那么出现1的个数由两部分组成，当前位减一出现1的个数和从当前位开始总共有多少个数,因为如果当前位是1，那么前面只要存在一个数，那么1的总数就会加一。
详细见代码：

#include <iostream>
#include <string>
#include <cstring>
#include <iomanip>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <map>
#include <stack>
#include <queue>
using namespace std;
#define ll long long
#define maxn 20
#define angel 0x3f3f3f3f
#define eps 1e-9
const int mod=1e9+7;
struct maths
{
    ll cnt;//出现1的个数
    ll sum;//总共多少个数
}dp[maxn];
int tmp[maxn];
maths dfs(int len,int flag)
{
    if(len==0)
    {
        maths element;
        element.sum=1;element.cnt=0;
        return element;
    }
    if(!flag&&dp[len].sum!=-1)
        return dp[len];
    int maxx=flag?tmp[len]:9;
    maths now;
    now.cnt=now.sum=0;
    for(int i=0;i<=maxx;i++)
    {
        maths next=dfs(len-1,i==maxx&&flag);
        if(i==1)
            now.cnt=now.cnt+next.cnt+next.sum;
        else
            now.cnt=now.cnt+next.cnt;
        now.sum+=next.sum;
    }
    if(!flag)
        dp[len]=now;
    return now;
}
ll solve(int n)
{
    int len=0;
    while(n>0)
    {
        tmp[++len]=n%10;
        n/=10;
    }
    return dfs(len,1).cnt;
}
int main()
{
    int n;
    scanf("%d",&n);
    memset(dp,-1,sizeof(dp));
    printf("%lld\n",solve(n));
    return 0;
}