1140 Look-and-say Sequence (20 point(s))
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
经验总结:
这一题,由于不知道40次之后会有多大,所以使用string来进行操作比较稳妥,难度不大,就不多说啦~
AC代码
#include <cstdio>
#include <string>
#include <iostream>
using namespace std;
int n;
int main()
{
string str,temp;
cin>>str>>n;
for(int i=0;i<n-1;++i)
{
temp.clear();
char str1[20];
int num=1;
for(int j=1;j<str.size();++j)
{
if(str[j]==str[j-1])
++num;
else
{
temp+=str[j-1];
sprintf(str1,"%d",num);
temp+=str1;
num=1;
}
}
temp+=str[str.size()-1];
sprintf(str1,"%d",num);
temp+=str1;
str=temp;
}
printf("%s",str.c_str());
return 0;
}