[PAT] 1140 Look-and-say Sequence（20 分）

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

题意：
D是0到9中除1以外的数字。第n+1个数字是第n个数字的一种描述。举个例子，第二个数表示第一个数中只有一个D，因此为D1；第二个数由一个D（D1）和一个1（11）组成，因此第三个数是D111；第四个数是D113，它由一个D（D1），三个1（13）组成；下一个是D11231。
你被要求计算第n个数对于给定的D。

思路：
用一个vector接收。每一次从头开始遍历。
注意一下vector数组结尾时的数据也要进行保存。

题解：

 1 #include<cstdlib>
 2 #include<cstdio>
 3 #include<vector>
 4 using namespace std;
 5 int main() {
 6     int d, n;
 7     scanf("%d %d", &d, &n);
 8     vector<int> num;
 9     num.push_back(d);
10     for (int i = 1; i < n; i++) {
11         vector<int> temp;
12         int v = num[0];
13         int cnt = 0;
14         for (int j = 0; j < num.size(); j++) {
15             if (num[j] == v) cnt++;
16             else{
17                 temp.push_back(v);
18                 temp.push_back(cnt);
19                 cnt = 1;
20                 v = num[j];
21             }
22             //如果到达结尾时，要最后一组数据进行push
23             if (j == num.size() - 1) {
24                 temp.push_back(v);
25                 temp.push_back(cnt);
26             }
27         }
28         num = temp;
29     }
30     for (int i = 0; i < num.size(); i++) {
31         printf("%d", num[i]);
32     }
33     return 0;
34 }