问题描述：
1140. Look-and-say Sequence (20)
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 8Sample Output:
1123123111
纪念一下3月18日PAT满分×1
这一题关键是要理解题意，理解后就不难了。。。
AC代码：
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#include<bits/stdc++.h>
using namespace std;
int main()
{
// freopen("data1.txt","r",stdin);
ios::sync_with_stdio(false);
vector<int> v;
int n,m;
cin>>n>>m;
m--;
v.push_back(n);
for(;m--;)
{
vector<int> v0;
int v00=v[0];
int ii=1;
for(int i=1;i<v.size();i++)
{
if(v[i]!=v00)
{
v0.push_back(v00);
v0.push_back(ii);
v00=v[i];
ii=1;
}
else
ii++;
}
v0.push_back(v00);
v0.push_back(ii);
v=v0;
}
for(auto i:v)
cout<<i;
return 0;
}
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