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

題意：
D是0到9中除1以外的數(shù)字。第n+1個(gè)數(shù)字是第n個(gè)數(shù)字的一種描述。舉個(gè)例子，第二個(gè)數(shù)表示第一個(gè)數(shù)中只有一個(gè)D，因此為D1；第二個(gè)數(shù)由一個(gè)D（D1）和一個(gè)1（11）組成，因此第三個(gè)數(shù)是D111；第四個(gè)數(shù)是D113，它由一個(gè)D（D1），三個(gè)1（13）組成；下一個(gè)是D11231。
你被要求計(jì)算第n個(gè)數(shù)對(duì)于給定的D。

思路：
用一個(gè)vector接收。每一次從頭開始遍歷。
注意一下vector數(shù)組結(jié)尾時(shí)的數(shù)據(jù)也要進(jìn)行保存。

題解：

#include<cstdlib>
#include<cstdio>
#include<vector>
using namespace std;
int main() {
    int d, n;
    scanf("%d %d", &d, &n);
    vector<int> num;
    num.push_back(d);
    for (int i = 1; i < n; i++) {
        vector<int> temp;
        int v = num[0];
        int cnt = 0;
        for (int j = 0; j < num.size(); j++) {
            if (num[j] == v) cnt++;
            else{
                temp.push_back(v);
                temp.push_back(cnt);
                cnt = 1;
                v = num[j];
            }
            //如果到達(dá)結(jié)尾時(shí)，要最后一組數(shù)據(jù)進(jìn)行push
            if (j == num.size() - 1) {
                temp.push_back(v);
                temp.push_back(cnt);
            }
        }
        num = temp;
    }
    for (int i = 0; i < num.size(); i++) {
        printf("%d", num[i]);
    }
    return 0;
}