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:给出一个数字d,再给出迭代次数n,让你按要求输出d在迭代n次后的数字,规则如下:
设第一项为d,
第二项为d1,其中d为上一项的第一位数字,1为d连续出现的次数。
第三项为d111,其中d1,表示d出现了一次,11表示1出现了1次。
第四项为d113,规则同理等等。
思路:使用两个while循环,第一个while控制第n项,第二个while用来处理字符串s,具体处理上,使用i指向当前数字,用cnt统计s[i]出现的次数,并把s[i]和次数cnt加入到答案字符串ans的后面,重复此操作直到字符串结束。
参考代码:
#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
using namespace std;
int main()
{
int n;
string s;
cin>>s>>n;
while(--n){
string ans="";
int i=0;
while(i<s.size()){
int cnt=1,j=i+1; //cnt统计s[i]出现的次数
while(j<s.size()&&s[j]==s[i]) {
j++;
cnt++;
}
ans+=s[i];
ans+=to_string((long long)cnt);
i=j;
}
s=ans;
}
cout<<s<<endl;
return 0;
}