【数位dp+状压】XHXJ 's LIS

题目

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

分析

题目的大概意思就是让你统计在给定区间内，符合要求的数的个数。 一个数如果它的各个数位的最长上升子序列长度为k，那么它就是符合要求的。 这题分为三个点。
1.首先这个题符合区间减法，我们只需要求出0~l-1和0~r的合法数的个数，再做减法即可。 
2.对LIS的处理我们采用状态压缩来处理
LIS状压：这个数的二进制的第i个1的位置表示当前序列长度为i的LIS最后一位最小是多少。这样1的个数就是LIS的长度
状态更新： 假设现在状态为0100100110，最长序列为1 4 7 8，如果我们下一个dp位的值为6，那么长度为3的上升子序列就由原来的1 4 7，变为1 4 6。 相应的我们更新后状态为0100101010，相当于6把7在二进制数上替换了。 然后来一遍深搜结束。

代码

#include <bits/stdc++.h>
using namespace std;
long long dp[25][1<<10][11];
int k,bit[25];
int ne(int x,int s){
    for (int i=x;i<10;++i)
    if (s&1<<i) return (s^(1<<i)|(1<<x));
    return s|(1<<x);
}
int num(int s){
    int ret=0;
    while (s){
        if (s&1)
        ret++;
        s>>=1;
    }
    return ret;
}
long long dfs (int pos,int s,bool e,bool z){
    if (pos==-1) return num(s)==k;
    if (!e&&dp[pos][s][k]!=-1) return dp[pos][s][k];
    long long ans=0;
    int endd=e?bit[pos]:9;
    for (int i=0;i<=endd;++i)
    ans+=dfs(pos-1,(z&&i==0)?0:ne(i,s),e&&i==endd,z&&(i==0));
    if (!e) dp[pos][s][k]=ans;
    return ans;
}
long long ca(long long n){
    int len=0;
    while (n){
        bit[len++]=n%10;
        n/=10;
    }
    return dfs(len-1,0,1,1);
}
int main(){
    int t;
    long long l,r;
    memset(dp,-1,sizeof dp);
    scanf("%d",&t);
    int casee=0;
    while (t--){
        scanf("%I64d%I64d%d",&l,&r,&k);
        printf("Case #%d: ",++casee);
        printf("%I64d\n",ca(r)-ca(l-1));
    }
    return 0;
}