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
程序:
#include <cstdio>
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int d,n;
scanf("%d%d",&d,&n);
vector<int> v;
v.push_back(d);
for(int i = 1; i < n; i++)
{
vector<int> temp;
temp.push_back(v[0]);
int count = 1;
for(int j = 1; j < v.size(); j++)
{
if(v[j] == v[j-1])
count++;
else
{
temp.push_back(count);
temp.push_back(v[j]);
count = 1;
}
}
temp.push_back(count);
v.clear();
v = temp;
temp.clear();
}
for(int i = 0; i < v.size(); i++)
{
printf("%d",v[i]);
}
return 0;
}