HDU 3943 數位dp+二分

K-th Nya Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 3155    Accepted Submission(s): 1021

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

 

 

Sample Output

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

 

 

Author

hzhua

 題意：

求區間內第k個有且僅有x個4和y個7的數

代碼：

//注意細節
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int t,x,y,bit[30];
ll f[30][25][25],P,Q;
ll dfs(int pos,int xx,int yy,int limit)
{
    if(pos==0) return (xx==0&&yy==0);
    if(xx<0||yy<0) return 0;
    if(!limit&&f[pos][xx][yy]!=-1) return f[pos][xx][yy];
    int max_b=limit?bit[pos]:9;
    ll ans=0;
    for(int i=0;i<=max_b;i++){
        ans+=dfs(pos-1,xx-(i==4),yy-(i==7),limit&&(i==max_b));
    }
    if(!limit) f[pos][xx][yy]=ans;
    return ans;
}
ll solve(ll O)
{
    int pos=0;
    while(O){
        bit[++pos]=O%10;
        O/=10;
    }
    return dfs(pos,x,y,1);
}
int main()
{
    //freopen("in.txt","r",stdin);
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++){
        printf("Case #%d:\n",cas);
        memset(f,-1,sizeof(f));
        scanf("%lld%lld%d%d",&P,&Q,&x,&y);
        ll tmp1=solve(P);
        ll tmp2=solve(Q);
        int n;ll k;
        scanf("%d",&n);
        while(n--){
            scanf("%lld",&k);
            if(tmp2-tmp1<k) { puts("Nya!");continue; }
            ll l=P+1,r=Q,ans;
            while(l<=r){
                ll mid=(l+r)>>1;
                ll tmp=solve(mid);
                if(tmp-tmp1>=k) { ans=mid;r=mid-1; }
                else l=mid+1;
            }
            printf("%lld\n",ans);
        }
    }
    return 0;
}