HDU 4352 XHXJ's LIS(数位DP+状压)

#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。

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).

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.

  
1
123 321 2

  
Case #1: 139 

peterae86@UESTC03

2012 Multi-University Training Contest 6

int turn(int s,int x)//状态转移 对状态s加入数字 x
{for (int i = x;i < 10;++i)//类似LIS中的更新末位元素使其变小{if (s&(1<<i)) //将这个比 x 大的数字 i 用 x这个更小的数去更新return (s^(1<<i))|(1<<x);//s^(1<<i)是把 i 这一位变成0， |(1<<x)则将 x 这一位变成 1}//如果 x 比该 LIS序列的最后一个元素都大则可以加入到后面去return s|(1<<x);//直接将这一位变成 1
}

/* Description: 最长上升子序列  定义d[k]:长度为k的上升序列最末元素(终点元素) //注意d中元素是单调递增的 初始化:len = 1,d[1]=a[1], 然后对于a[i]: if a[i] > d[len]   len++,d[len] = a[i] else  from d[1] to d[len]找到一个j 满足 d[j-1]<a[i]<d[j] 更新d[j] = a[i]  
*/ 

#define mem(a,x) memset(a,x,sizeof(a))
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<stdlib.h>
#include<cctype>
#include<string>
using namespace std;
typedef long long ll;
ll dp[22][1<<10][11];
/*dp[i][j][k]的解释：i 表示当前进行到的数位 （好像基本上数位DP的第一维都是保存的这个）j 是一个压缩的状态，该状态记录的是一个LIS序列，该状态中的 1 的个数即为最长上升子序列的长度k 最长上升子序列的长度为 kdp[i][j][k]满足 i,j,k 的数的个数
*/
int K;
int turn(int s,int x)//状态转移 对状态s加入数字 x
{for (int i = x;i < 10;++i)//类似LIS中的更新末位元素使其变小{if (s&(1<<i)) //将这个比 x 大的数字 i 用 x这个更小的数去更新return (s^(1<<i))|(1<<x);//s^(1<<i)是把 i 这一位变成0， |(1<<x)则将 x 这一位变成 1}//如果 x 比该 LIS序列的最后一个元素都大则可以加入到后面去return s|(1<<x);//直接将这一位变成 1
}
int getlen(int s)//返回该状态表示的最长上升子序列的长度
{int t = 0;while (s){if (s&1) ++t;s>>=1;}return t;
}
int dig[22];//数的每一个数位
/*关于 0 的标记：搜长度为4的时候会搜到 0000搜长度为3的时候会搜到 000但实际上0000和000是同一个数用 zero 记录前一位是不是0
*/
ll dfs(int len,int s,int up,int zero)//当前进行的数位，状态，是否为上界，前一位是否为0
{if (len == 0) return getlen(s) == K;if (!up&&dp[len][s][K]!=-1) return dp[len][s][K];ll ans = 0;int n = 9;if (up) n = dig[len];for (int i = 0;i <= n;++i){if (i == 0){/*0 不能为首位 否则会重复计算当前面有数字的时候就可以加入0*/if (zero) ans += dfs(len-1,0,up&&(i==n),zero&&(i==0));else ans += dfs(len-1,turn(s,i),up&&(i==n),zero&&(i==0));}else ans += dfs(len-1,turn(s,i),up&&(i==n),zero&&(i==0));}if (!up) dp[len][s][K] = ans;return ans;
}
ll cal(ll x)
{int len = 0;while (x){dig[++len] = x%10;x/=10;}return dfs(len,0,1,1);
}
int main()
{int T;scanf("%d",&T);int kas = 0;mem(dp,-1);while (T--){ll l,r;scanf("%I64d %I64d %d",&l,&r,&K);printf("Case #%d: %I64d\n",++kas,cal(r)-cal(l-1));}return 0;
}