Information
- ID
- 49
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 7
- Tags
- # Submissions
- 604
- Accepted
- 151
- Uploaded By
一道简单的动态规划问题,设到第i节台阶的方案数为dp[i],则dp[i] = dp[i-1] + dp[i-2];
#include<bits/stdc++.h>
using namespace std;
long long n,dp[75];
int main(){
cin>>n;
dp[0] = dp[1] = 1;
for(int i = 2;i<=n;i++){
dp[i] = dp[i-1]+dp[i-2];
}
cout<<dp[n];
return 0;
}
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