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.

InputThe 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).OutputFor 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

Sample Output

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

題目大概：

給定p,q,x,y,找出在p和q之間的   含有x個4和y個7的數中   第k大個數。

思路：

這個題和昨天做的一道題，類似，都是最簡單的數位dp基礎上加上二分就好了。

數位dp部分就不說了，誰都會。

在p和q之間二分，利用查找出來的符合條件的數量減去q的符合條件的數量就是第n大的數，利用二分找出第k大個數，就好了，也是基礎二分。

主要是精度控制。

程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
LL  INF=0xfffffffffLL;
int a[65];
LL dp[65][65][65];
LL x,y;
LL sove(int pos,int q4,int q7,int limit)
{
    if(q4>x||q7>y)return 0;
    if(pos==-1)return (q4==x)&&(q7==y);
    if(!limit&&dp[pos][q4][q7]!=-1)return dp[pos][q4][q7];
    int end=limit?a[pos]:9;
    LL ans=0;
    for(int i=0;i<=end;i++)
    {
        if(i==4)ans+=sove(pos-1,q4+1,q7,limit&&i==end);
        else if(i==7)
        {
            ans+=sove(pos-1,q4,q7+1,limit&&i==end);
        }
        else ans+=sove(pos-1,q4,q7,limit&&i==end);
    }
    if(!limit)dp[pos][q4][q7]=ans;
    return ans;
}
LL go(LL x)
{
    int pos=0;
    while(x)
    {
        a[pos++]=x%10;
        x/=10;
    }
    return sove(pos-1,0,0,1);
}
int main()
{   int t;
    LL n;
    scanf("%d",&t);
    for(int j=1;j<=t;j++)
    {   printf("Case #%d:\n",j);
        memset(dp,-1,sizeof(dp));
        LL p,q;
        int n;
        scanf("%I64d%I64d%I64d%I64d",&p,&q,&x,&y);
        scanf("%d",&n);
        LL w1=go(q);
        LL w2=go(p);
        while(n--)
        {

        LL k;
        scanf("%I64d",&k);
        if(k>w1-w2)
        {
            printf("Nya!\n");
            continue;
        }
        LL l=p,r=q,mid;
        while(l<=r)
        {
            mid=(l+r)/2;
            if(go(mid)-w2<k)
            {
                l=mid+1;
            }
            else
            {
                r=mid-1;
            }

        }
       printf("%I64d\n",l);
        }

    }


    return 0;
}