【HDU3943】【K-th Nya Number】【数位+二分找位置】

K-th Nya Number
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 2493    Accepted Submission(s): 770


Problem Description
Arcueid likes nya number very much.
A nya number is the number which has exactly X fours and Y sevens(If X=2 and Y=3 , 172441277 and 47770142 are nya numbers.But 14777 is not a nya number ,because it has only 1 four).
Now, Arcueid wants to know the K-th nya number which is greater than P and not greater than Q.
 
Input
The first line contains a positive integer T (T<=100), indicates there are T test cases.
The second line contains 4 non-negative integers: P,Q,X and Y separated by spaces.
( 0<=X+Y<=20 , 0< P<=Q <2^63)
The third line contains an integer N(1<=N<=100).
Then here comes N queries.
Each of them contains an integer K_i (0<K_i <2^63). 
Output
For each test case, display its case number and then print N lines.
For each query, output a line contains an integer number, representing the K_i-th nya number in (P,Q].
If there is no such number,please output "Nya!"(without the quotes).
 
Sample Input
1
38 400 1 1
10
1
2
3
4
5
6
7
8
9
10

Case #1:
47
74
147
174
247
274
347
374
Nya!
Nya!

#include <iostream>#include <cstring>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <cstdlib>#include <cstdio>#include <algorithm>using namespace std;    typedef long long ll;int T;ll P,Q,X,Y;int N;int bits[70];ll dp[70][70][70];ll dfs(int len,int c4,int c7,bool flag){	if(c4 > X || c7 > Y) return 0;	if(len <= 0) return c4 == X && c7 == Y;	if(!flag && dp[len][c4][c7] != -1) return dp[len][c4][c7];	int end = flag ? bits[len] : 9;	ll res  = 0;	for(int i=0;i<=end;i++)	{		if(i == 4)		res += dfs(len-1,c4+1,c7,flag && i == end);		else if(i == 7)		res += dfs(len-1,c4,c7+1,flag && i == end);		else		res += dfs(len-1,c4,c7,flag && i == end);	}	return flag ? res : dp[len][c4][c7] = res;}ll solve(ll x){	int c = 0;	ll tt = x;	while(tt > 0)	{		bits[++c] = tt % 10;		tt /= 10;	}	return dfs(c,0,0,true);}int main(){	int C = 1;    scanf("%d",&T);	while(T --)	{		printf("Case #%d:\n",C++);		memset(dp,-1,sizeof(dp));		scanf("%I64d%I64d%I64d%I64d",&P,&Q,&X,&Y);		ll dn = solve(P);		ll up = solve(Q);		scanf("%d",&N);		while(N--)		{			ll m;			scanf("%I64d",&m);			ll l = P + 1;		    ll r = Q;			if(m > up -dn) 			{				printf("Nya!\n");				continue;			}			while(l < r)			{				ll mid = (l + r) / 2;				ll s = solve(mid);				s - dn >= m ? (r = mid) : (l = mid + 1);			}				printf("%I64d\n",l);		}	}    return 0;}