PAT 甲级 1140 Look-and-say Sequence (20分)
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 that 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 11), therefore the 3rd number is D111
; or since the 4th number is D113
, it consists of one D
, two 1's, and one 3, so the next number must be D11231
. This definition works for D
= 1 as well. Now you are 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 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
题目大意:第n+1个串是第n个串的描述, 若第n个串是11122234,那么第n+1个串就是13233141,即1有3个、2有3个、3有1个、4有1个。
样例解释:暂无
【AC代码】:
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int main() 5 { 6 int n; string s; cin >> s >> n; 7 for(int i = 2; i <= n; i++) 8 { 9 string k; int top = 1; 10 for(int j = 1; j <= s.length(); j++) 11 { 12 if(s[j] == s[j-1]) top++; 13 else{ 14 k += s[j-1] + to_string(top); 15 top = 1; 16 } 17 } 18 s = k; 19 } 20 cout << s; 21 return 0; 22 }