3 条题解

  • 1
    @ 2026-7-29 11:52:41

    #include<bits/stdc++.h> #define endl '\n' using namespace std; const int N=1e5+5,INF=0x3f3f3f3f,MOD=1e9+7; const int dx[]={1,-1,0,0},dy[]={0,0,1,-1}; typedef long long LL; LL n,minn=INF,maxx=-INF,x,dp1,dp2; int main(){ ios::sync_with_stdio(false); cin.tie(nullptr);cout.tie(nullptr); cin>>n; for(int i=1;i<=n;i++){ cin>>x; LL a=dp1,b=dp2; dp1 = max({x,xa,xb}),dp2 = min({x,xa,xb}); maxx = max(maxx,dp1),minn = min(minn,dp2); } cout<<maxx<<endl<<minn; return 0; }

    • 1
      @ 2025-11-2 18:16:48

      冲击最小空间!

      #include<bits/stdc++.h>
      #define endl '\n'
      using namespace std;
      const int N=1e5+5,INF=0x3f3f3f3f,MOD=1e9+7;
      const int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
      typedef long long LL;
      LL n,minn=INF,maxx=-INF,x,dp1,dp2;
      int main(){
      	ios::sync_with_stdio(false);
      	cin.tie(nullptr);cout.tie(nullptr);
      	cin>>n;
      	for(int i=1;i<=n;i++){
      		cin>>x;
      		LL a=dp1,b=dp2;
      		dp1 = max({x,x*a,x*b}),dp2 = min({x,x*a,x*b});
      		maxx = max(maxx,dp1),minn = min(minn,dp2);
      	}
      	cout<<maxx<<endl<<minn;
      	return 0;
      }
      
      
      • -2
        @ 2025-10-26 20:14:31

        蠢方法(暴力)

        #include <bits/stdc++.h>
        using namespace std;
        const int N = 1e7 + 10;
        const int INF = 0x3f3f3f3f;
        int n;
        long long a[N],dpm[N],dpn[N],nmax = -INF,nmin = INF;
        int main(){
        	cin >> n;
        	for(int i = 1;i <= n;i++){
        		cin >> a[i];
        	}
        	dpn[0] = dpm[0] = 1;
        	for(int i = 1;i <= n;i++){
        		dpm[i] = a[i];
        		dpm[i] = max(a[i],dpm[i - 1] * a[i]);
        		dpn[i] = a[i];
        		dpn[i] = min(a[i],dpn[i - 1] * a[i]);
        		dpm[i] = max(dpm[i],dpn[i - 1] * a[i]);
        		dpn[i] = min(dpn[i],dpm[i - 1] * a[i]);
        		//printf("dpm[%d] = %d\n",i,dpm[i]);
        		//printf("dpn[%d] = %d\n",i,dpn[i]);
        		nmax = max(nmax,dpm[i]);
        		nmin = min(nmin,dpn[i]);
        	}
        	cout << nmax << endl << nmin;
        	return 0;
        }    
        
        
        • 1

        信息

        ID
        1764
        时间
        1000ms
        内存
        256MiB
        难度
        8
        标签
        递交数
        132
        已通过
        24
        上传者