HDU 5504 GT and sequence 模拟

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5504

思路：模拟

代码：

 #include<stdio.h>//------杭电5504
 #include<algorithm>
 #include<math.h>
 #include<iostream>
 using namespace std;
 int comp(const void*a, const void*b)
 {
     return *(int*)a - *(int*)b;
 }
 int main()
 {
     int t;
     scanf("%d", &t);
     while (t)
     {
         __int64 *a;
         int n;
         int count1 = ;//记录负数
         int count2 = ;//记录0
         int count3 = ;//记录正数
         int i;
         __int64 result = ;
         __int64 maxx = -0x3f3f3f3f;
         scanf("%d", &n);
         a = (__int64*)malloc(n*sizeof(__int64));
         for (i = ; i < n; i++)
         {
             scanf("%I64d", (a + i));
             if (a[i] < )
             {
                 count1++;
                 maxx = max(maxx, a[i]);
             }
             else if (a[i]>)
                 count3++;
             else
                 count2++;
         }
         qsort(a, n, sizeof(__int64), comp);//对数组排序
         if (count1 == )//没有负数
         {
             if (count3 != )
             {
                 if (count2 == )
                     i = ;
                 else
                     i = count2;
                 for (i; i < n; i++)
                 {
                     result *= a[i];
                 }
             }
             else if (count3 == )//无正数
             {
                 result = ;
             }
         }
         else if (count1 == )//只有一个负数
         {
             if (count3 != )
             {
                 if (count2 == )
                     i = ;
                 else
                     i = count2 + count1;
                 for (i; i < n; i++)
                 {
                     result *= a[i];
                 }
             }
             else if (count3 == )//无正数
             {
                 if (count2 == )
                     result *= a[];
                 else
                     result = ;
             }
         }
         else if (count1 %  == )//偶数个负数
         {
             for (i = ; i < count1; i++)
             {
                 result *= a[i];
             }
             if (count3 != )
             {
                 if (count2 == )
                     i = count1;
                 else
                     i = count2 + count1;
                 for (i; i < n; i++)
                 {
                     result *= a[i];
                 }
             }
             else if (count3 == )//无正数，值保留
             {
                 result = result;
             }
         }
         else if (count1 %  != )//奇数个负数
         {
             for (i = ; i < count1-; i++)
             {
                 result *= a[i];
             }
             if (count3 != )
             {
                 if (count2 == )
                     i = count1;
                 else
                     i = count2 + count1;
                 for (i; i < n; i++)
                 {
                     result *= a[i];
                 }
             }
             else if (count3 == )//无正数，值保留
             {
                 result = result;
             }
         }
         printf("%I64d\n", result);
         t--;
     }
     return ;
 }