P2602 [ZJOI2010] digital count counter & P4999 & P1239 mathematical operations annoying

P2602 [ZJOI2010] digital count

answer

DFS nausea digital DP

For this question, we can find a number of a number

I.e., respectively, count the number of interval [L, R] i arise internal digital (0 <= i <= 9)

 

DFS is the only record:

Current fill the first of several pos

How many times have appeared a total sum k

Target number k

Whether top border limit

Are all leading zeros qdl

 

dp[pos][sum]：

> No top sector without leading zeros,

  Current pos fill in the first place, the target figure sum contribution times when the answers appear to produce a total of

> Since the sum will be taken up to the same number and pos, the array size so as to open and like pos

 

Record sum when there are two cases:

(1) k! = 0 filled directly to see whether the target figures like digital

(2) k = 0 <1> are all leading zeros in front, but the number is not filled 0

                     <2> 0 is filled to the last digit, that is 0000000, to be considered at this time 0 appear once

                     <3> to rest the situation is the record number 0 appears

 

 

Code

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<queue>

using namespace std;

typedef long long ll;

inline ll read()
{
    ll ans=0;
    char last=' ',ch=getchar();
    while(ch<'0'||ch>'9') last=ch,ch=getchar();
    while(ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
    if(last=='-') ans=-ans;
    return ans;
}

ll a,b;
ll c[20],len;
ll dp[20][20];

ll dfs(ll pos,ll sum,ll k,bool limit,bool qdl)
{
    if(pos<=0) return sum;
    if(!limit&&!qdl&&dp[pos][sum]!=-1) return dp[pos][sum];
    ll ans=0;
    ll up=limit?c[pos]:9;
    for(int i=0;i<=up;i++)
        ans+=dfs(pos-1,sum+(k==0?(!qdl&&i==0)||(qdl&&i==0&&pos==0):(i==k)),k,limit&&(i==up),qdl&&(i==0));
    if(!limit&&!qdl) dp[pos][sum]=ans;
    return ans;
}

ll sum(ll x,ll k)
{
    memset(c,0,sizeof(c));len=0;
    memset(dp,-1,sizeof(dp));
    while(x)
    {
        c[++len]=x%10;
        x/=10;
    }
    return dfs(len,0,k,1,1);
}

int main()
{
    a=read();b=read();
    for(int i=0;i<=9;i++)
      printf("%lld ",sum(b,i)-sum(a-1,i));
    
    return 0;
}

 

 

 

---

 

Double experience (simpler than the first one)

P1239 counter

answer

That is, only one is enough enough of a

 

Code

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<queue>

using namespace std;

typedef long long ll;

inline ll read()
{
    ll ans=0;
    char last=' ',ch=getchar();
    while(ch<'0'||ch>'9') last=ch,ch=getchar();
    while(ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
    if(last=='-') ans=-ans;
    return ans;
}

const ll mod=1e9+7;
ll a,b,T,ans=0;
ll c[30],len;
ll dp[30][30];

ll dfs(ll pos,ll sum,ll k,bool limit,bool qdl)
{
    if(pos<=0) return sum;
    if(!limit&&!qdl&&dp[pos][sum]!=-1) return dp[pos][sum];
    ll ans=0;
    ll up=limit?c[pos]:9;
    for(int i=0;i<=up;i++)
        ans+=dfs(pos-1,sum+(k==0?(!qdl&&i==0)||(qdl&&i==0&&pos==0):(i==k)),k,limit&&(i==up),qdl&&(i==0));
    if(!limit&&!qdl) dp[pos][sum]=ans;
    return ans;
}

ll sum(ll x,ll k)
{
    memset(c,0,sizeof(c));len=0;
    memset(dp,-1,sizeof(dp));
    while(x)
    {
        c[++len]=x%10;
        x/=10;
    }
    return dfs(len,0,k,1,1);
}

int main()
{
    a=read();
    for(int i=0;i<=9;i++)
       printf("%lld\n",sum(a,i));    
    return 0;
}

 

 

 

---

 

Three times the experience

P4999 annoying math homework

answer

Take a look at the question: I want AC spicy! ! !

The reality is 90pt

As Ha! ! ! ! ! !

 

Modulo of pot, take a few more

 

 

I was too naive

 

 

Code

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<queue>

using namespace std;

typedef long long ll;

inline ll read()
{
    ll ans=0;
    char last=' ',ch=getchar();
    while(ch<'0'||ch>'9') last=ch,ch=getchar();
    while(ch>='0'&&ch<='9') ans=ans*10+ch-'0',ch=getchar();
    if(last=='-') ans=-ans;
    return ans;
}

const ll mod=1e9+7;
ll a,b,T,ans=0;
ll c[20],len;
ll dp[20][20];

ll dfs(ll pos,ll sum,ll k,bool limit,bool qdl)
{
    if(pos<=0) return sum;
    if(!limit&&!qdl&&dp[pos][sum]!=-1) return dp[pos][sum];
    ll ans=0;
    ll up=limit?c[pos]:9;
    for(int i=0;i<=up;i++)
        ans+=dfs(pos-1,sum+(k==0?(!qdl&&i==0)||(qdl&&i==0&&pos==0):(i==k)),k,limit&&(i==up),qdl&&(i==0));
    if(!limit&&!qdl) dp[pos][sum]=ans;
    return ans;
}

ll sum(ll x,ll k)
{
    memset(c,0,sizeof(c));len=0;
    memset(dp,-1,sizeof(dp));
    while(x)
    {
        c[++len]=x%10;
        x/=10;
    }
    return dfs(len,0,k,1,1);
}

int main()
{
    T=read();
    while(T--)
    {
        ans=0;
        a=read();b=read();
        for(int i=1;i<=9;i++)
        {
            ll tmp=((sum(b,i)-sum(a-1,i))%mod+mod)%mod; 

            ans=((ans+i*tmp%mod+mod)%mod+mod)%mod;
        }
           
        printf("%lld\n",ans%mod);
    }
    
    return 0;
}