XHXJ's LIS

XHXJ's LIS

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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<2⁶³-1 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：一开始想当前位置，当前出现了哪些数，当前LIS是多少为同一个状态，不过不对；

　　　考虑LIS是怎么求的，有nlogn做法，更新出现过的第一个大于等于自己的值；

　　　所以dp[i][j][k]三维分别表示当前位置，当前LIS里的数，剩余剩余所需长度；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <bitset>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define piii pair<int,pair<ll,ll> >
#define sys system("pause")
const int maxn=1e5+;
const int N=5e4+;
const int M=N**;
using namespace std;
inline ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
inline ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
inline void umax(ll &p,ll q){if(p<q)p=q;}
inline void umin(ll &p,ll q){if(p>q)p=q;}
int n,m,k,t,num[],pos,cas;
ll dp[][<<][],p,q;
ll dfs(int pos,int x,int y,int z,int v)
{
    if(pos<)return y==;
    if(y<)return ;
    if(z&&v&&dp[pos][x][y]!=-)return dp[pos][x][y];
    int now=z?:num[pos],i,j;
    ll ret=;
    rep(i,,now)
    {
        rep(j,i,)if(x>>j&)break;
        if(!i&&!v)ret+=dfs(pos-,x,y,z||i<num[pos],v||i);
        else if(j==)ret+=dfs(pos-,x|(<<i),y-,z||i<num[pos],v||i);
        else ret+=dfs(pos-,x^(<<j)^(<<i),y,z||i<num[pos],v||i);
    }
    return z&&v?dp[pos][x][y]=ret:ret;
}
ll gao(ll x)
{
    pos=;
    while(x)num[pos++]=x%,x/=;
    return dfs(pos-,,k,,);
}
int main()
{
    int i,j;
    memset(dp,-,sizeof(dp));
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%d",&p,&q,&k);
        printf("Case #%d: %lld\n",++cas,gao(q)-gao(p-));
    }
    return ;
}