模板 - 数位dp

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

int a[20];
ll dp[20][MAXS1][MAXS2];
ll dfs(int pos,int s1,int s2,bool lead,bool limit) {
    if(pos==-1) {
        return ?;
    }
    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++) {
        int ns1=op1(s1);
        int ns2=op2(s2);
        ans+=dfs(pos-1,ns1,ns2,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 ?;

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

    return dfs(pos-1,INITS1,INITS2,true,true);
}

int main() {
    memset(dp,-1,sizeof(dp));

    ll le,ri;
    while(~scanf("%lld%lld",&le,&ri)) {
        printf("%lld\n",solve(ri)-solve(le-1));
    }
}

来源：https://www.cnblogs.com/Inko/p/11393887.html