【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=20;
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]!=-1&&!limit)||deep==0)
{
qp.num=NUM[deep];
qp.ans=f[deep];
return qp;
}
int k=limit?a[deep]:9;
int sum=0;long long ans=0;
for(int i=0;i<=k;i++)
{
qp=dfs(deep-1,limit&&i==k);
if(i==1)ans+=qp.num;
ans+=qp.ans;
sum+=qp.num;
}
if(!NUM[deep]&&!limit)NUM[deep]=sum;
if(f[deep]==-1&&!limit)f[deep]=ans;
qp.ans=ans;qp.num=sum;
return qp;
}
int main()
{
scanf("%d",&n);
int k=n;int N=0;
while(k>0)
{
N++;
a
=k%10;
k/=10;
}
memset(f,-1,sizeof(f));
NUM[0]=1;f[0]=0;
printf("%lld",dfs(N,1).ans);
return 0;
}