题意：求t个n位的2进制数中，有多少个没有连续的1.
dp0[x]表示以0开头长度为x的这类01串的个数，dp1[x]表示以1开头长度为x的这类01串的个数
dp0[x] = dp0[ x-1 ] + dp1[ x-1 ], dp0[1] = 1;
dp1[x] = dp0[ x-1 ] , dp1[1] = 1;
合并 dp0[x] + dp1[x] = dp0[x-1] + dp1[x-1] + dp0[x-1]
又dp0[x-1] = dp0[x-2] + dp1[x-2]
得：
ans[i] = ans[i-1] + ans[i-2]
由于第i位可任选0或1：
选1时，i-1位选0，i-2位选1的合法
选0时，i-1位选1，i-2位选0的合法
# include <iostream>
# include <algorithm>
using namespace std;
const int _ = 0;
long long ans[50] = {0};
int t, n;
int main()
{
int i;
ans[1] = 2LL, ans[2] = 3LL;
for ( i = 3; i < 50; ++i ) ans[i] = ans[i-1] + ans[i-2];
cin >> t;
for ( i = 1; i <= t; ++i )
{
cin >> n;
cout << "Scenario #" << i <<":" << endl;
cout << ans[n] << endl << endl;
}
return 0>_<0;
}