最大子序列和

找了一道题目HDU - 1231，写法有很多。

我的最优战绩109ms，我真的不晓得Leaderboard里62ms，甚至还有15ms的大佬们是怎么做到的orz！

Code（dp递推未优化）

#include 
#define INF 0x3f3f3f3f

using namespace std;

const int maxn = 10050;

int dp[maxn];
int a[maxn];
int n;
int ans;
int Left, Right;

int findLeft()
{
    int index;
    for (index = Right; index >= 1; index--)
    {
        if (dp[index - 1] < 0) break;
    }
    return index;
}

int main()
{
    while (cin >> n && n)
    {
        for (int i = 0; i < n; i++) cin >> a[i];

        dp[0] = a[0];
        ans = dp[0];

        Left = Right = 0;

        for (int i = 1; i < n; i++)
        {
            dp[i] = max(0, dp[i - 1]) + a[i];

            if (dp[i] > ans)
            {
                Right = i;
                ans = dp[i];
            }
        }

        Left = findLeft();
        if (ans >= 0) cout << ans << " " << a[Left] << " " << a[Right] << endl;
        else cout << 0 << " " << a[0] << " " << a[n - 1] << endl;
    }

    return 0;
}

本段代码测试时间：702ms

再优化 （dp递推优化）

#include 
#define INF 0x3f3f3f3f

using namespace std;

const int maxn = 10050;

int dp[maxn];
int a[maxn];
int n;
int ans;
int Left, Right;

int findLeft()
{
    int index;
    for (index = Right; index >= 1; index--)
    {
        if (dp[index - 1] < 0) break;
    }
    return index;
}

int main()
{
    while (scanf("%d", &n) != EOF && n)
    {
        scanf("%d", &a[0]);

        dp[0] = a[0];
        ans = dp[0];

        Left = Right = 0;

        for (int i = 1; i < n; i++)
        {
            scanf("%d", &a[i]);
            dp[i] = max(0, dp[i - 1]) + a[i];

            if (dp[i] > ans)
            {
                Right = i;
                ans = dp[i];
            }
        }

        Left = findLeft();
        if (ans >= 0) cout << ans << " " << a[Left] << " " << a[Right] << endl;
        else cout << 0 << " " << a[0] << " " << a[n - 1] << endl;
    }

    return 0;
}

耗时缩短至124ms

 

Code（O(n)优化，省去了找左边界）

#include 
#define INF 0x3f3f3f3f

using namespace std;

const int maxn = 10050;

int a[maxn];
int n;
int ans;
int Left, Right;


int main()
{
    while (scanf("%d", &n) != EOF && n)
    {
        scanf("%d", &a[0]);

        ans = a[0];

        Left = Right = 0;

        int sum = a[0];
        int flag = Left;

        for (int i = 1; i < n; i++)
        {
            scanf("%d", &a[i]);

            if (sum < 0)
            {
                sum = a[i];
                flag = i;
            }
            else sum += a[i];

            if (sum > ans)
            {
                ans = sum;
                Right = i;
                Left = flag;
            }
        }

        if (ans >= 0) cout << ans << " " << a[Left] << " " << a[Right] << endl;
        else cout << 0 << " " << a[0] << " " << a[n - 1] << endl;
    }

    return 0;
}

耗时仍然为124ms

 

Code（继续优化，考虑到数据问题，我们将全部为负数的情况加以特判，时间成功缩减至109ms）

#include 
#define INF 0x3f3f3f3f

using namespace std;

const int maxn = 10001;

int a[maxn];
int n;

int main()
{
    while (scanf("%d", &n) != EOF && n)
    {
        int plot = 0;

        for (int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            if (a[i] >= 0) plot = 1;
        }

        if (!plot) printf("0 %d %d\n", a[0], a[n - 1]);
        else
        {
            int ans = a[0];
            int Left = 0, Right = 0;
            int sum = a[0];
            int flag = Left;

            for (int i = 1; i < n; i++)
            {
                if (sum < 0)
                {
                    sum = a[i];
                    flag = i;
                }
                else sum += a[i];

                if (sum > ans)
                {
                    ans = sum;
                    Right = i;
                    Left = flag;
                }
            }
            printf("%d %d %d\n", ans, a[Left], a[Right]);
        }
    }

    return 0;
}