【51NOD】数字1的数量

【算法】数位DP

【题解】数位dp总结 之 从入门到模板

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=;
int n,a[maxn],NUM[maxn];
long long f[maxn];
struct cyc{int num/*数字数*/;long long ans/*1的数量*/;}qp;
cyc dfs(int deep,bool limit)//返回1的个数
{
    if((f[deep]!=-&&!limit)||deep==)
    {
        qp.num=NUM[deep];
        qp.ans=f[deep];
        return qp;
    }
    int k=limit?a[deep]:;
    int sum=;long long ans=;
    for(int i=;i<=k;i++)
    {
        qp=dfs(deep-,limit&&i==k);
        if(i==)ans+=qp.num;
        ans+=qp.ans;
        sum+=qp.num;
    }
    if(!NUM[deep]&&!limit)NUM[deep]=sum;
    if(f[deep]==-&&!limit)f[deep]=ans;
    qp.ans=ans;qp.num=sum;
    return qp;
}
int main()
{
    scanf("%d",&n);
    int k=n;int N=;
    while(k>)
    {
        N++;
        a[N]=k%;
        k/=;
    }
    memset(f,-,sizeof(f));
    NUM[]=;f[]=;
    printf("%lld",dfs(N,).ans);
    return ;
}