ALGORITHM_01

Day1

O($n^3$)

#include<cstdio>
#include<cmath>
using namespace std;
#define MAX_N 110
int n,a[MAX_N];
int max_length =0;
void solve()
{
    for(int i = 0;i < n;i++)
        for(int j = i+1;j<n;j++)
            for(int k = j+1;k<n;k++)
            {
                int length = a[i] + a[j] + a[k];
                int max_a = max(a[i],max(a[j],a[k]));
                int residue = length - max_a;
                if(residue > max_a && length > max_length)
                {
                    max_length = length;
                }
            }
}
int main()
{
    scanf("%d",&n);
    for(int i = 0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    solve();
    printf("%d",max_length);
    system("pause");
    return 0;
}

O($n\ast logn$)

最优解必定是排完序后连续的三个数

反正法：

1. 假设$a,b,c$是最优解，存在$a<a^{.}<b<b^{.}<c$
2. 因为$a+b>c$,所以一定有$a^{.}+b>c$,因此$a^{.},b,c$优于$a,b,c$,矛盾，故不存在$a^{.}$
3. $b^{.}$同理不存在

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
#define MAX_N 110
int n,a[MAX_N];
int max_length =0;
void solve()
{
    for(int i = n-1;i >=2;i--)
    {
        int length = a[i] + a[i-1] + a[i-2];
        int residue = length - a[i];
        if(residue > a[i] && length > max_length)
        {
            max_length = length;
        }
    }
}
int main()
{
    scanf("%d",&n);
    for(int i = 0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    sort(a,a+n);
    solve();
    printf("%d",max_length);
    system("pause");
    return 0;
}