PAT Advanced 1140. Look-and-say Sequence (20)

Problem Description:

1140. Look-and-say Sequence time limit (ms)
Memory Limit 65536 KB
Code length limit 16000 B
Program Standard author CHEN, Yue

Look-and-say sequence is a sequence of integers as the following:

D, D1, D111, D113, D11231, D112213111, ...

where D is in [0, 9] except 1. The (n+1) St number is a kind of description of the nth number. For example, the 2nd number means this there is one D in the 1st number, and hence it is D1; The 2nd number consists of one D (corresponding to D1) and one 1 (corresponding to one), therefore the 3rd number is D111; or since the 4th number is D113, it consists of one D, and both 1 ' s, and one 3, so the next number must be D11231. This definition works for D = 1 as well. Now is supposed to calculate the Nth number in a look-and-say sequence of a given digit D.

Input Specification:

Each input file contains one test case, which gives D (in [0, 9]) and a positive an integer N (<=40), separated by a space .

Output Specification:

Print in a line the Nth number in a look-and-say sequence of D. Sample Input:

1 8

Sample Output:

1123123111

Remember March 18 Pat out of X1

The key to this problem is to understand the test instructions, after understanding it is not difficult ...

AC Code:

1 2 3 4 5 6 7 8 9 30 of each of the above. 37 off

#include <bits/stdc++.h> using namespace std; int main () { // freopen ("Data1.txt", "R", stdin); Ios::sync_with_stdio (false); Vector<int> v; int n,m; cin>>n>>m; m--; V.push_back (n); for (; m--;) { vector<int> v0; int v00=v[0]; int ii=1; for (int i=1;i<v.size (); i++) { if (v[i]!=v00) { v0.push_back (v00); V0.push_back (ii); V00=v[i]; Ii=1; } else ii++; } V0.push_back (v00); V0.push_back (ii); v=v0; } for (auto i:v) cout<<i; return 0; }