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 =  1000000007;
ll l,r;
int k,lbit[20],rbit[20],used[20][210][1010];
struct node{
ll num,left;//num能发的人数、left前一子树剩余和
node(ll num = 0, ll left = 0) : num(num), left(left) {}
node operator += (node b)
{
num += b.num;
left = b.left;
return *this;
}
}dp[20][210][1010];
node dfs(int i,int sum,int left,int le,int re){
if(i==0){
if(sum+left>=k)
return node(1,0);
return node(0,sum+left);
}
if(used[i][sum][left]&&!le&&!re)
return dp[i][sum][left];
int ll=le?lbit[i]:0;
int rr=re?rbit[i]:9;
node a(0,left);
for(int v=ll;v<=rr;++v){
a+=dfs(i-1,sum+v,a.left,le&&(v==ll),re&&(v==rr));
}
if(!le&&!re){
dp[i][sum][left]=a;
used[i][sum][left]=1;
}
return a;
}
void solve(){
memset(used,0,sizeof(used));
memset(dp,0,sizeof(dp));
int len=0;
while(r){
rbit[++len]=r%10;
r/=10;
lbit[len]=l%10;
l/=10;
}
node t=dfs(len,0,0,1,1);
printf("%I64d\n",t.num);
}
int main()
{
scanf("%I64d%I64d%d",&l,&r,&k);
solve();
return 0;
}