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
#include<bits/stdc++.h> using namespace std; typedef long long ll; int main(){ string s; cin >> s; int n; cin >> n;n--; while(n--){ string temp = ""; for(int i=0;i < s.size();){ int j = i; while(s[i] == s[j]&&j < s.size()){ j++; } temp += s[i]+to_string(j-i); i = j; } s = temp; } cout << s; return 0; }
——这题意我也是醉了,完全没说清楚,还把人误导用map,看了题解才知道啥意思。