NTHU OJ - 7665 PD - Cake wants to win

The following program is my ACcepted code for NTHU-7492 .
It's a for everybody to learn and discuss.
If there is any mistake or comment, please let me know.  :D

此乃NTHU 7492 的AC code!

歡迎一同討論學習，如有錯誤與任何建議請留言 : )

點這裡看題目 Click here to see this Problem!

先把卡片的數字累加!
讓card[i] 代表 第一張卡片加到第i張卡片的值
之後只要決定 "開頭" 跟 "結尾" 的index
就可以用 card[結尾]-card[開頭] = 從 "開頭" 到 "結尾" 的合
所以只要用 O(N^2) 得時間複雜度

//This program is for NTHU 7665 PD - Cake wants to win
//題目來源 Problem link: http://acm.cs.nthu.edu.tw/problem.php?pid=7665

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX 2001

int main()
{
    int N, K, tmp, ans;
    int card[MAX];//card[i] is sum from 1st card to ith card.
    bool findAns = false;
 
    while(scanf("%d %d", &N, &K) != EOF)
    {
        memset(card, 0, sizeof(card));
        findAns = false;
     
        for(int i = 1; i <= N; i++)
        {
            scanf("%d", &tmp);
            card[i] = card[i - 1] + tmp;
        }
     
        for(int i = 0; i < N; i++)
        {
            for(int j = i + 1; j <= N; j++)
            {
                tmp = card[j] - card[i];
             
                if(tmp > K && !findAns)
                {
                    findAns = true;
                    ans = tmp;
                }
                else if(findAns)
                {
                    if(tmp > K && tmp < ans)
                        ans = tmp;
                }
            }
        }
        if(!findAns)
        {
            // if we can't find answer, but K < 0.
            //Then we just don't choose any card and put 0!
            if(K < 0)
                printf("0\n");
            else
                printf("Cake will lose QAQ\n");
        }
     
        if(findAns)
        {
            //because K < 0, if answer < 0, then "this answer" is better than 0.
            if(K < 0 && ans >= 0)
                printf("0\n");
            else
                printf("%d\n", ans);
        }
    }
    return 0;
}

Please feel free to use it after adding this blog as an reference. (http://autekroy.blogspot.tw) If there is any mistake or comment, please let me know. :D 

歡迎使用與分享任何內容，但請記得標示此部落格為出處。(http://autekroy.blogspot.tw/) 如果有發現任何的錯誤與建議請留言或跟我連絡。 : )