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 }