Problem #
Source: [PAT 1140]
Description #
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
Solution #
- 题意 给你一个数字D和次数n,执行n次f操作,f操作是计算当前数字从前往后每个连续数字出现的次数,如1经行f操作为11,1121134进行f操作为1221123141
- 思路 以字符串的形式进行f操作
Code #
#include <iostream>
#include <algorithm>
#include <string>
#define maxsize 202
using namespace std;
int m, n;
string s, tmp;
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(0);
cin >> s >> m;
for (int i = 0; i < m-1; i++)
{
tmp = s[0];
n = 1;
int j;
for (j = 1; j < s.size(); j++)
{
if (s[j] == s[j - 1])
n++;
else
{
tmp += ('0' + n);
tmp += s[j];
n = 1;
}
}
if (j == s.size())
tmp += ('0' + n);
s = tmp;
}
cout << s << endl;
return 0;
}