1140 Look-and-say Sequence

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

迭代的方法读字符串，把每次读取

string s

的结果存入

string tmp

中，读完后

s = tmp

再继续读

string s

即可。

因为最多的迭代次数也只有 40 次，耗时 16ms；空间复杂度 O(n)

#include <iostream>
#include <string>

using namespace std;

int main() {
int n;
string s;
cin >> s >> n;					//首个字符串是单数字的
for(int i = 1; i < n; ++i) {	//注意循环从 1 开始
string tmp;
for(int j = 0; j < s.size();++j) {
tmp += s[j];
int cnt = 1;
while(s[j+1] == s[j]) { 	//此次结束循环的条件
j++;					//j 自增在循环内部执行
cnt++;
}							//跳转到下一个不相同字符通过 for 循环的 ++j 执行
tmp += to_string(cnt);
}
s = tmp;
}
cout << s;
return 0;
}

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