SRM 498 div2

Jun 07, 2016 pm 03:15 PM

250pt: 题意:就是黑板上有三个数,有n次选择的机会,每次选择哪个数可以得到和那个数相同的分数,然后黑板上的该数减一。求n次选择最多能得到多少分。(注意分数减到0后就不在减)。 思路:水题啊,每次选择时排序即可,选择最大的。 代码: int getMaximu

250pt:

题意:就是黑板上有三个数,有n次选择的机会,每次选择哪个数可以得到和那个数相同的分数,然后黑板上的该数减一。求n次选择最多能得到多少分。(注意分数减到0后就不在减)。

思路:水题啊,每次选择时排序即可,选择最大的。

代码:

int getMaximumPoints(int A, int B, int C, int N)
        {
            int i,j,k;
			int sum = 0;
			int num[3] = {A,B,C};
			for(int i = 1;i = 1)
			  num[2]--;
			}
			return sum;
        }
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500pt:

题意:有这样一种序列,s[0] s[a+1] > s[a+2] > s[a+3] > ... >s[b] = s[b+1] = s[b+2] = ... = s[c] s[d+1] > s[d+2] > ... >s[n-1],而且小于的值都相等,也就是说s[0] 到 s[a] 是一个等差数列,s[a]到s[b]是一个等差数列,s[c] 到s[d] 是一个等差数列,s[d] 到 s[n-1] 是一个等差数列。判断所给的序列中是否满足这样的条件。对应输出“YES” 和“NO” 即可。

思路:其实就是模拟题了,不过细节问题需要注意,考虑清楚即可。

代码:

string isValid(vector <int> seq)
        {
            int i,j,k;
			string ss;
			int flag = true;
			int len = seq.size();
			if(len = len)
				  break;
			  while(seq[i+1] - seq[i] == x &amp;&amp; x &gt; 0){
			     i++;
			  }
			  a = i;
			  x = seq[i+1] - seq[i];
			  while(seq[i+1] - seq[i] == x &amp;&amp; x  0){
			    i++;
			  }
			  d = i;
			  x = seq[i+1] - seq[i];
			  while(seq[i+1] - seq[i] == x &amp;&amp; x <br>
<br>



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