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
#include<iostream> #include<cstdio> #include<string> using namespace std; int hashTB[1000] = {0}; int main(){ string ss = ""; int D, N; cin >> D >> N; ss = ss + (char)('0' + D); for(int i = 0; i < N - 1; i++){ char pre = ss[0]; int cnt = 1; string ss2 = ""; for(int j = 1; j < ss.length(); j++){ if(ss[j] == pre){ cnt++; }else{ ss2 += pre; ss2 += (char)(cnt + '0'); pre = ss[j]; cnt = 1; } } ss2 += pre; ss2 += (char)(cnt + '0'); ss = ss2; } cout << ss; cin >> N; return 0; }
总结:
1、题意:题意不好理解时可以根据标题来猜意思。这题的意思让你看并且读出一个序列, 如一个序列是18882111,则可以读为1个1,3个8,1个2,3个1,再将其写成位在前个数描述在后的序列:11,83,21,13。再对这个新序列进行描述。