不连续1的子串 题解:long long
#include <iostream>
#include <string>
#include <cstring>
using namespace std;
int main()
{
int N;
cin>>N;
long long total[21]={1};
total[1]=2;
total[2]=3;
int x=2;
if(N>2)
while(x++!=N)
total[x]=total[x-1]+total[x-2];
cout<<total[N]<<endl;
return 0;
}
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