最新代码qwq
```
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
long long l,r,len,up,ans;
long long c[20],dp[20][20][20][20][20];
long long Dp(long long k,long long last,bool limit,bool qdl,long long x,long long sum) //现在填到了第k位,上一位是几,是否顶上界,是否含有前导零
{
if(k==0) return sum;
if(dp[k][last][limit][qdl][x]!=-1) return dp[k][last][limit][qdl][x];
if(limit) up=c[k];
else up=9;
long long res=0;
for(long long i=0;i<=up;i++)
{
if(qdl&&i==0) res+=Dp(k-1,i,limit&&i==up,1,x,sum);
else res+=Dp(k-1,i,limit&&i==up,0,x,sum+(i==x));
}
dp[k][last][limit][qdl][x]=res;
return res;
}
long long solve(long long x,long long y)
{
if(x==0) return 0;
len=0;
while(x)
{
c[++len]=x%10;
x/=10;
}
memset(dp,-1,sizeof(dp));
return Dp(len,0,1,1,y,0);
}
int main()
{
scanf("%lld%lld",&l,&r);
for(long long i=0;i<=9;i++) printf("%lld ",solve(r,i)-solve(l-1,i));
return 0;
}
```
by 暗ざ之殇 @ 2019-11-03 11:13:24
更新代码(哭了)
```
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
long long dp[100][100],c[100];
long long up,res,len,l,r;
long long Dp(int k,bool limit,bool qdl,long long sum,int x)
{
if(k==0) return sum;
if(!limit&&!qdl&&dp[k][sum]!=-1) return dp[k][sum];
if(limit) up=c[k];
else up=9;
res=0;
for(int i=0;i<=up;i++)
{
if(qdl&&i==0) res+=Dp(k-1,limit&&i==up,1,sum,x);
else res+=Dp(k-1,limit&&i==up,0,sum+(i==x),x);
}
if(!limit&&!qdl) dp[k][sum]=res;
return res;
}
long long solve(long long x,int y)
{
len=0;
while(x)
{
c[++len]=x%10;
x/=10;
}
memset(dp,-1,sizeof(-1));
return Dp(len,1,1,0,y);
}
int main()
{
scanf("%lld%lld",&l,&r);
for(int i=0;i<=9;i++) printf("%lld ",solve(r,i)-solve(l-1,i));
return 0;
}
```
by 暗ざ之殇 @ 2019-11-03 11:51:32
找到原因了:up和res必须要在函数里定义?为什么呀?
```
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
long long dp[100][100],c[100];
long long len,l,r;
long long Dp(int k,bool limit,bool qdl,long long sum,int x)
{
if(k==0) return sum;
if(!limit&&!qdl&&dp[k][sum]!=-1) return dp[k][sum];
long long up,res=0; //在函数里定义
if(limit) up=c[k];
else up=9;
res=0;
for(int i=0;i<=up;i++)
{
if(qdl&&i==0) res+=Dp(k-1,limit&&i==up,1,sum,x);
else res+=Dp(k-1,limit&&i==up,0,sum+(i==x),x);
}
if(!limit&&!qdl) dp[k][sum]=res;
return res;
}
long long solve(long long x,int y)
{
len=0;
while(x)
{
c[++len]=x%10;
x/=10;
}
memset(dp,-1,sizeof(dp));
return Dp(len,1,1,0,y);
}
int main()
{
scanf("%lld%lld",&l,&r);
for(int i=0;i<=9;i++) printf("%lld ",solve(r,i)-solve(l-1,i));
return 0;
}
```
by 暗ざ之殇 @ 2019-11-03 15:13:36