HDU 3943 K-th Nya Number

K-th Nya Number

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on HDU. Original ID: 3943
64-bit integer IO format: %I64d      Java class name: Main

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!

Source

2011 Multi-University Training Contest 11 - Host by UESTC

 

解题：数位dp+二分

#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int maxn = 22;
LL dp[maxn][maxn][maxn];
int bt[maxn],x,y;
LL dfs(int len,int a,int b,bool flag){
    if(len == -1) return a == x && y == b;
    if(!flag && dp[len][a][b] != -1) return dp[len][a][b];
    int u = flag?bt[len]:9;
    LL ans = 0;
    for(int i = 0; i <= u; ++i){
        if(i == 4) ans += dfs(len-1,a + 1,b,flag&&i == u);
        else if(i == 7) ans += dfs(len-1,a,b + 1,flag&&i==u);
        else ans += dfs(len-1,a,b,flag&&i==u);
    }
    if(!flag) dp[len][a][b] = ans;
    return ans;
}
LL solve(LL x){
    int cnt = 0;
    while(x){
        bt[cnt++] = x%10;
        x /= 10;
    }
    return dfs(cnt - 1,0,0,true);
}
int main(){
    int kase,m,cs = 1;
    scanf("%d",&kase);
    while(kase--){
        LL P,Q,K;
        scanf("%I64d%I64d%d%d",&P,&Q,&x,&y);
        memset(dp,-1,sizeof dp);
        scanf("%d",&m);
        printf("Case #%d:
",cs++);
        LL tmp = solve(P);
        while(m--){
            scanf("%I64d",&K);
            K += tmp;
            LL low = P + 1,high = Q,ans = -1;
            while(low <= high){
                LL mid = (low + high)>>1;
                if(solve(mid) >= K){
                    ans = mid;
                    high = mid - 1;
                }else low = mid + 1;
            }
            if(ans == -1) puts("Nya!");
            else printf("%I64d
",ans);
        }
    }
    return 0;
}