hdu4352 XHXJ's LIS 数位DP+状态压缩

XHXJ's LIS

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1084    Accepted Submission(s): 427

Problem Description

#define xhxj (Xin Hang senior sister(学姐))
If you do not know xhxj, then carefully reading the entire description is very important.
As the strongest fighting force in UESTC, xhxj grew up in Jintang, a border town of Chengdu.
Like many god cattles, xhxj has a legendary life:
2010.04, had not yet begun to learn the algorithm, xhxj won the second prize in the university contest. And in this fall, xhxj got one gold medal and one silver medal of regional contest. In the next year's summer, xhxj was invited to Beijing to attend the astar onsite. A few months later, xhxj got two gold medals and was also qualified for world's final. However, xhxj was defeated by zhymaoiing in the competition that determined who would go to the world's final(there is only one team for every university to send to the world's final) .Now, xhxj is much more stronger than ever，and she will go to the dreaming country to compete in TCO final.
As you see, xhxj always keeps a short hair(reasons unknown), so she looks like a boy( I will not tell you she is actually a lovely girl), wearing yellow T-shirt. When she is not talking, her round face feels very lovely, attracting others to touch her face gently。Unlike God Luo's, another UESTC god cattle who has cool and noble charm, xhxj is quite approachable, lively, clever. On the other hand,xhxj is very sensitive to the beautiful properties, "this problem has a very good properties"，she always said that after ACing a very hard problem. She often helps in finding solutions, even though she is not good at the problems of that type.
Xhxj loves many games such as，Dota, ocg, mahjong, Starcraft 2, Diablo 3.etc，if you can beat her in any game above, you will get her admire and become a god cattle. She is very concerned with her younger schoolfellows, if she saw someone on a DOTA platform, she would say: "Why do not you go to improve your programming skill". When she receives sincere compliments from others, she would say modestly: "Please don’t flatter at me.(Please don't black)."As she will graduate after no more than one year, xhxj also wants to fall in love. However, the man in her dreams has not yet appeared, so she now prefers girls.
Another hobby of xhxj is yy(speculation) some magical problems to discover the special properties. For example, when she see a number, she would think whether the digits of a number are strictly increasing. If you consider the number as a string and can get a longest strictly increasing subsequence the length of which is equal to k, the power of this number is k.. It is very simple to determine a single number’s power, but is it also easy to solve this problem with the numbers within an interval? xhxj has a little tired，she want a god cattle to help her solve this problem,the problem is: Determine how many numbers have the power value k in [L，R] in O(1)time.
For the first one to solve this problem，xhxj will upgrade 20 favorability rate。

 

Input

First a integer T(T<=10000),then T lines follow, every line has three positive integer L,R,K.(
0<L<=R<2 ⁶³-1 and 1<=K<=10).

 

Output

For each query, print "Case #t: ans" in a line, in which t is the number of the test case starting from 1 and ans is the answer.

 

Sample Input

   
   
   
   

    
    
    
    1
123 321 2
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 139 
   
   
   
   

  区间[L,R]有多少个数满足最长上升子序列长度为K。

  这道题思路类似于nlogn最长上升子序列的思路。数位DP，状态s保存搜索到当前位的最长上升子序列（状态压缩，如果有i这个数，那么s的1<<i这一位就是1）。如果当前位取n，得到新s的方法是看当前最长上升子序列中有没有存在大于等于n的数，如果有，找到最小的那个大于等于n的数，用n替换，这样长度不变，并且可能会减小序列中的数，算是贪心的思想吧。如果没有大于等于n的，就把n加进去。到最后判断s中1的个数是否为K。

  注意前面都是0的情况是不能把0加到s里的，所以搜索的时候要判断前导0。

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;

const int MAXN=30;
const int MAXM=12;
const int INF=0x3f3f3f3f;
const int MOD=2520;

int T,K;
int bit[MAXN];
LL L,R;
LL dp[MAXN][1<<10][MAXM];

int cal(int s){
    int ret=0;
    for(int i=0;i<=9;i++) if(s&(1<<i)){
        ret++;
    }
    return ret;
}

int get_news(int s,int n){
    for(int i=n;i<=9;i++) if(s&(1<<i)){
        return (s^(1<<i))|(1<<n);
    }
    return s|(1<<n);
}

LL dfs(int cur,int s,int e,int z){
    if(cur<0) return cal(s)==K;
    if(!e&&!z&&dp[cur][s][K]!=-1) return dp[cur][s][K];
    int end=e?bit[cur]:9;
    LL ans=0;
    for(int i=0;i<=end;i++){
        if(z&&!i) ans+=dfs(cur-1,s,e&&i==end,1);
        else ans+=dfs(cur-1,get_news(s,i),e&&i==end,0);
    }
    if(!e&&!z) dp[cur][s][K]=ans;
    return ans;
}

LL solve(LL n){
    int len=0;
    while(n){
        bit[len++]=n%10;
        n/=10;
    }
    return dfs(len-1,0,1,1);
}

int main(){
    freopen("in.txt","r",stdin);
    scanf("%d",&T);
    int cas=0;
    memset(dp,-1,sizeof(dp));
    while(T--){
        scanf("%I64d%I64d%d",&L,&R,&K);
        printf("Case #%d: %I64d\n",++cas,solve(R)-solve(L-1));
    }
    return 0;
}