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：后一个字符串是前一个字符串的描述。
思路：模拟
代码：
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <sstream>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define mem(a,n) memset(a,n,sizeof(a))
#define lowbit(i) ((i)&(-i))
typedef long long ll;
typedef unsigned long long ull;
const ll INF=0x3f3f3f3f;
const double eps = 1e-6;
const int N = 2e2+5;
int main() {
int n;
string str;
cin>>str>>n;
for(int k,i=1; i<n; i++) {
string tmp;
for(int j=0; j<str.length(); j=k) {
for(k=j; k<str.length(); k++) {
if(str[k]!=str[j]) {
break;
}
}
tmp+=to_string(str[j]-'0')+to_string(k-j);
}
str=tmp;
}
cout<<str;
return 0;
}
