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
首先看看样例是怎么变成1123123111的
1
11
12
1121
122111
112213
12221131
1123123111
本题要点:
- string 只能用 cin cout 输入输出,不能使用 scanf 与 printf
- t = t +…与 t+=… 后者效率较高
- to_string()将数字转为字符串
#include <iostream>
#include <string>
using namespace std;
int main() {
int N, j;
string s;
cin >> s >> N;
for(int cnt = 1;cnt < N; cnt++){
string t;
for(int i = 0; i < s.length(); i = j){
for(j = i; j < s.length() && s[j] == s[i]; j++);
t += s[i] + to_string(j - i); //如果使用t=t+...会超时
}
s = t;
}
cout << s;
return 0;
}
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