数位DP POJ

Posted 2020-09-26

Apocalypse Someday

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 1870		Accepted: 902

Description

The number 666 is considered to be the occult “number of the beast” and is a well used number in all major apocalypse themed blockbuster movies. However the number 666 can’t always be used in the script so numbers such as 1666 are used instead. Let us call the numbers containing at least three contiguous sixes beastly numbers. The first few beastly numbers are 666, 1666, 2666, 3666, 4666, 5666…

Given a 1-based index n, your program should return the nth beastly number.

Input

The first line contains the number of test cases T (T ≤ 1,000).

Each of the following T lines contains an integer n (1 ≤ n ≤ 50,000,000) as a test case.

Output

For each test case, your program should output the nth beastly number.

Sample Input

3
2
3
187

Sample Output

1666
2666
66666

Source

POJ Monthly--2007.03.04, Ikki, adapted from TCHS SRM 2 ApocalypseSomeday

 

 

 

我不明白我为什么要写这个题，好难……

lyd的题解 OTZ

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 int t,n,m,k;
 7 long long f[21][4];
 8 void work(){
 9     f[0][0]=1;
10     for(int i=0;i<20;i++){
11         for(int j=0;j<3;j++){
12             f[i+1][j+1]+=f[i][j];
13             f[i+1][0]+=f[i][j]*9;
14         }
15         f[i+1][3]+=f[i][3]*10;
16     }
17 }
18 int main(){
19     work();
20     scanf("%d",&t);
21     while(t){
22         t--;
23         scanf("%d",&n);
24         for(m=3;f[m][3]<n;m++);
25         k=0;
26         for(int i=m;i;i--){
27             for(int j=0;j<=9;j++){
28                 long long tmp=f[i-1][3];
29                 if(k==3||j==6)
30                     for(int p=max(0,3-k-(j==6));p<3;p++)
31                         tmp+=f[i-1][p];
32                 if(tmp<n)  n-=tmp;
33                 else{
34                     if(k<3&&j==6) k++;
35                     if(k<3&&j!=6) k=0;
36                     printf("%d",j);
37                     break;
38                 }
39             }
40         }
41         printf("\\n");
42     }
43     return 0;
44 }