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  • PAT甲级:1136 A Delayed Palindrome (20分)

    PAT甲级:1136 A Delayed Palindrome (20分)

    题干

    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
    

    思路

    一道经典的滑动窗口题目。而我遇到了一个坑点。

    • ans = ans + d[q] + to_string(p - q);
    • ans += d[q] + to_string(p - q);

    咋一看这两句是同一个意思,但是第一句在PAT测试点3是超时的。

    教训就是,别闲着没事乱写,老老实实写+= ,明明就更简单是不是?

    具体原因不太清楚,反正遇上了,就试试换成+= 看对不对?

    code

    #include <iostream>
    #include <string>
    using namespace std;
    int main(){
    	int n = 0;
    	string d;
    	d.resize(1);
    	scanf("%s%d", &d[0], &n);
    	for(int i = 1; i < n; i++){
    		string ans;
    		int p = 1, q = 0;
    		d = d + "@";
    		while(p < d.size()){
    			if(d[p] != d[q]){
    				//ans = ans + d[q] + to_string(p - q);
    				ans += d[q] + to_string(p - q);
    				q = p;
    			}
    			p++;
    		}
    		d = ans;
    	}
    	printf("%s", d.c_str());
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/cell-coder/p/12840648.html
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