SGU 390-Tickets（数位dp）

题意：有标号l-r的票，要给路人发，当给的票的编号的各数位的总和（可能一个人多张票）不小k时，才开始发给下一个人，求能发多少人。

分析：这个题挺难想的，参考了一下题解，dp[i][sum][left] 长度i 当前数位和sum  前一子树剩余的和

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cctype>
#include <complex>
#include <cassert>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
#define lson l,m,rt<<1
#define pi acos(-1.0)
#define rson m+1,r,rt<<11
#define All 1,N,1
#define read freopen("in.txt", "r", stdin)
const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF= 0x7ffffff;
const int mod =  ;
ll l,r;
int k,lbit[],rbit[],used[][][];
struct node{
    ll num,left;//num能发的人数、left前一子树剩余和
    node(ll num = , ll left = ) : num(num), left(left) {}
    node operator += (node b)
    {
        num += b.num;
        left = b.left;
        return *this;
    }
}dp[][][];
node dfs(int i,int sum,int left,int le,int re){
    if(i==){
        if(sum+left>=k)
        return node(,);
        return node(,sum+left);
    }
    if(used[i][sum][left]&&!le&&!re)
        return dp[i][sum][left];
    int ll=le?lbit[i]:;
    int rr=re?rbit[i]:;
    node a(,left);
    for(int v=ll;v<=rr;++v){
        a+=dfs(i-,sum+v,a.left,le&&(v==ll),re&&(v==rr));
    }
    if(!le&&!re){
        dp[i][sum][left]=a;
        used[i][sum][left]=;
    }
    return a;
}
void solve(){
    memset(used,,sizeof(used));
    memset(dp,,sizeof(dp));
    int len=;
    while(r){
        rbit[++len]=r%;
        r/=;
        lbit[len]=l%;
        l/=;
    }
    node t=dfs(len,,,,);
    printf("%I64d\n",t.num);
}
int main()
{
    scanf("%I64d%I64d%d",&l,&r,&k);
        solve();
return ;
}