SCUT - 289 - 小O的数字 - 数位dp

Posted 2022-01-13 yinku

https://scut.online/p/289
一个水到飞起的模板数位dp。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

bool notp[2000];

const int MAXS1=200;
const int MAXS2=2000;

int a[20];
ll dp[20][MAXS1][MAXS2];
ll dfs(int pos,int s1,int s2,bool lead,bool limit) 
    if(pos==-1) 
        if(notp[s1]||notp[s2])
            return 0;
        else
            return 1;
    
    if(!limit && !lead && dp[pos][s1][s2]!=-1)
        return dp[pos][s1][s2];
    int up=limit?a[pos]:9;
    ll ans=0;
    for(int i=0; i<=up; i++) 
        ans+=dfs(pos-1,s1+i,s2+i*i,lead && i==0,limit && i==a[pos]);
    
    if(!limit && !lead)
        dp[pos][s1][s2]=ans;
    return ans;


ll solve(ll x) 
    if(x<=0)
        return 0;

    int pos=0;
    while(x) 
        a[pos++]=x%10;
        x/=10;
    

    return dfs(pos-1,0,0,true,true);


int main() 
#ifdef Yinku
    freopen("Yinku.in","r",stdin);
#endif // Yinku
    memset(dp,-1,sizeof(dp));
    notp[0]=1;
    notp[1]=1;
    for(int i=2; i<2000; i++) 
        if(!notp[i]) 
            for(int j=i+i; j<2000; j+=i) 
                notp[j]=1;
            
        
    

    int T;
    scanf("%d",&T);
    ll le,ri;
    while(T--) 
        scanf("%lld%lld",&le,&ri);
        printf("%lld\n",solve(ri)-solve(le-1));