1 solutions

  • 0
    @ 2025-3-3 16:33:14

    C++ :

    #include<bits/stdc++.h>
    using namespace std;
    const int N=10010;
    int A[N],dp[N];
    int main() {
    	int n;
    	cin>>n;
    	for(int i=1; i<=n; i++) {
    		cin>>A[i];
    	}
    	int ans=-1;        //记录最大的dp[i]
    	for(int i=1; i<=n; i++) {     //按顺序计算出dp[i]的值
    		dp[i]=1;        //边界初始条件(即先假设每个元素自成一个子序列)
    		for(int j=1; j<i; j++) {
    			if(A[i]>A[j]&&(dp[j]+1>dp[i])) {
    				dp[i]=dp[j]+1;        //状态转移方程,用以更新dp[i]
    			}
    		}
    		ans=max(ans,dp[i]);
    	}
    	cout<<ans<<endl;
    	return 0;
    }
    
    
    • 1

    Information

    ID
    10544
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
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