2 条题解

  • 2
    @ 2024-4-9 17:23:40
    #include <iostream>
    #include <cmath>
    using namespace std;
    long long f[100002],a[100002] = {0};
    int main()
    {
    	long long pos = 0, neg = 0;
    	int n;
    	cin>>n;
    	cin>>a[1];
    	for(int i=2;i<=n;i++)
    	{
    		cin>>a[i];
    		f[i] = a[i]-a[i-1];
    		if(f[i]>0)
    			pos +=f[i];
    		else if(f[i]<0)
    			neg -=f[i];
    	}
    	cout<<max(pos,neg)<<endl;
    	cout<<abs(pos-neg)+1<<endl;
    	return 0;
    }
    
  • 2
    @ 2023-11-8 19:53:17
    #include <iostream>
    #include <cmath>
    using namespace std;
    long long f[100002],a[100002] = {0};
    int main()
    {
    	long long pos = 0, neg = 0;
    	int n;
    	cin>>n;
    	cin>>a[1];
    	for(int i=2;i<=n;i++)
    	{
    		cin>>a[i];
    		f[i] = a[i]-a[i-1];
    		if(f[i]>0)
    			pos +=f[i];
    		else if(f[i]<0)
    			neg -=f[i];
    	}
    	cout<<max(pos,neg)<<endl;
    	cout<<abs(pos-neg)+1<<endl;
    	return 0;
    }
    
    • 1

    信息

    ID
    12
    时间
    1000ms
    内存
    128MiB
    难度
    4
    标签
    递交数
    287
    已通过
    140
    上传者