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  • A. Puzzles CodeForces Round #196 (Div.2)

    题目的大意是,给你 m 个数字,让你从中选 n 个,使得选出的数字的极差最小。

      好吧,超级大水题。因为要极差最小,所以当然想到要排个序咯,然后去连续的 n 个数字,因为数据不大,所以排完序之后直接暴力就OK了。

    附AC代码:

       1: #include <stdio.h>
       2: #include <math.h>
       3: #include <iostream>
       4: #include <cstdarg>
       5: #include <algorithm>
       6: #include <string.h>
       7: #include <stdlib.h>
       8: #include <string>
       9: #include <list>
      10: #include <vector>
      11: #include <map>
      12: #define LL long long
      13: #define M(a) memset(a, 0, sizeof(a))
      14: using namespace std;
      15:  
      16: void Clean(int count, ...)
      17: {
      18:     va_list arg_ptr;
      19:     va_start (arg_ptr, count);
      20:     for (int i = 0; i < count; i++)
      21:         M(va_arg(arg_ptr, char*));
      22:     va_end(arg_ptr);
      23: }
      24:  
      25: int buf[59];
      26:  
      27: int main()
      28: {
      29:     int n, m;
      30:     while (~scanf("%d%d", &n, &m))
      31:     {
      32:         Clean(1, buf);
      33:         for (int i = 0; i < m; i++)
      34:             scanf("%d", &buf[i]);
      35:         sort(buf, buf + m);
      36:         int res = 999999;
      37:         for (int i = n - 1; i < m; i++)
      38:             res = min(res, (buf[i] - buf[i - n + 1]));
      39:         printf("%d
    ", res);
      40:     }
      41:     return 0;
      42: }
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  • 原文地址:https://www.cnblogs.com/wuhenqs/p/3267909.html
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