HDU 3943 K-th Nya Number（数位dp+二分）

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

138 400 1 11012345678910

 

Sample Output

Case #1:4774147174247274347374Nya!Nya!

 

Author

hzhua

 

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;#define fre(i,a,b)  for(i = a; i <b; i++)#define free(i,b,a) for(i = b; i >= a;i--)#define mem(t, v)   memset ((t) , v, sizeof(t))#define ssf(n)      scanf("%s", n)#define sf(n)       scanf("%d", &n)#define sff(a,b)    scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pf          printf#define bug         pf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 25ll dp[N][N][N];int bit[N];ll le,ri;int x,y,k;ll dfs(int pos,int prex,int prey,int bound){   if(pos==0) return prex==0&&prey==0;   if(prex<0||prey<0) return 0;   if(!bound&&dp[pos][prex][prey]!=-1) return dp[pos][prex][prey];   int up=bound ? bit[pos]:9;   ll temp=0;   int i;   fre(i,0,up+1)   {     temp+=dfs(pos-1,prex-(i==4),prey-(i==7),bound&&i==up);   }   if(!bound) dp[pos][prex][prey]=temp;   return temp;}ll fdd(ll n){   int len=0;   while(n)   {   bit[++len]=n%10;    n/=10;   }   return dfs(len,x,y,1);}void solve(){  ll prex=fdd(le);  ll prey=fdd(ri);  ll lle=le,rri=ri;  ll mid,ans;  k+=prex;  while(lle<=rri)  {  mid=(lle+rri)>>1;  if(fdd(mid)>=k){ans=mid;rri=mid-1;}    elselle=mid+1;  }   pf("%I64d\n",ans);}int main(){int i,j,t,ca=0;sf(t);mem(dp,-1);while(t--){pf("Case #%d:\n",++ca);scanf("%I64d%I64d",&le,&ri);sff(x,y);ll numle=fdd(le);ll numri=fdd(ri);//pf("%I64d %I64d\n",numle,numri);int m;sf(m);while(m--){sf(k);if(numle+k>numri)puts("Nya!");elsesolve();}}   return 0;}