hdu 1040

雪黛依梦 — Mon, 06 Sep 2010 13:19:00 GMT

#include
#include
using namespace std;

int main ()
{
    int a[1005];
    int t;
    scanf ( "%d", &t );
    while ( t -- )
    {
          int n;
          scanf ( "%d", &n );
          for ( int i = 0; i < n; i ++ )
          {
              scanf ("%d", &a[i]);
          }
          sort ( a, a + n );

          printf ("%d", a[0]);
          for ( int i = 1; i < n; i ++ )
          {
              printf (" %d",a[i]);
          }
          printf ("\n");
    }
   //system ("pause");
   return 0;
}

简单的排序题，直接利用sort默认的排序规则：升序排列

雪黛依梦 2010-09-06 21:19 发表评论

HDU 1280

雪黛依梦 — Sat, 28 Aug 2010 08:43:00 GMT

用 hash 做

Accepted

1280

0MS

292K

1236B

G++

//典型的hash: 用数组下标表示两两相加所得到的和，开辟一个满足题意的大小的数组 sum，
//这样下标由大到小输出m个就可以

#include <iostream>
#include <string>
using namespace std;

int main ()
{
    int a[3001];
    int sum[10001];
    int n, m;
    while ( scanf ("%d %d", &n, &m) != EOF )
    {
          memset ( a, 0, sizeof (a) );
          memset ( sum, 0, sizeof (sum) );
          for ( int i = 0; i < n; i ++ )
          {
              scanf ("%d", &a[i]);
          }

          int temp;
          for ( int i = 0; i < n; i ++ )
          {
              for ( int j = i + 1; j < n; j++ )
              {
                  temp = a[i] + a[j];
                  sum[temp] ++;
              }
          }

          int count = 0;      //输出前 m  个数
          for ( int i = 10001; i >= 0 ; i -- )
          {
              if ( sum[i] )
              {
                   for (int j = 0; j < sum[i]; j ++)
                   {
                       count ++;
                       count == 1 ? printf ("%d", i) : printf (" %d", i);
                         if ( count == m )
                             break;
                   }
                }
                      if ( count == m )
                      break;

          }

          printf ("\n");
    }

    // system ("pause");
     return 0;
}

所以改用sort排序后再输出

Accepted

1280

843MS

17888K

1304B

G++

//典型的hash: 用数组下标表示两两相加所得到的和，开辟一个满足题意的大小的数组 sum，
//这样下标由大到小输出m个就可以

#include
#include
using namespace std;

int sum[4500000];          //开全局数组才能过
bool cmp ( const int &a, const int &b )
{
     return a > b;
}

int main ()
{
    int a[3001];

    int n, m;
    while ( scanf ("%d %d", &n, &m) != EOF )
    {
          memset ( a, 0, sizeof (a) );
          memset ( sum, 0, sizeof (sum) );
          //输入处理
          for ( int i = 0; i < n; i ++ )
          {
              scanf ("%d", &a[i]);
          }

          //两数求和
          //int k = n * ( n - 1 ) / 2;
          int temp = 0;
          for ( int i = 0; i < n; i ++ )
          {
              for ( int j = i + 1; j < n; j ++ )
              {
                  sum[temp ++] = a[i] + a[j];
              }
          }

          sort ( sum, sum + temp, cmp );

          int count = 0;
          for ( int i = 0; i< temp; i ++ )
          {
              count ++;
              count == 1 ? printf ("%d", sum[i]): printf (" %d",sum[i]);

              if ( count == m )
              break;
          }
          printf ("\n");
    }

    // system ("pause");
     return 0;
}

雪黛依梦 2010-08-28 16:43 发表评论

C++博客-雪黛依梦-随笔分类-排序题

hdu 1040

HDU 1280