ZOJ 4025 King of Karaoke

最新推荐文章于 2019-10-10 16:27:34 发布

原创最新推荐文章于 2019-10-10 16:27:34 发布 · 462 阅读

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本文介绍了一种通过调整整数序列实现歌曲表演分数最大化的算法。针对两个长度相等的整数序列，通过选择合适的调整值使得序列尽可能匹配，以此来提高计分。文章提供了一个具体的C语言实现案例。

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题目:
It’s Karaoke time! DreamGrid is performing the song Powder Snow in the game King of Karaoke. The song performed by DreamGrid can be considered as an integer sequence D₁, D₂,……,D_n, and the standard version of the song can be considered as another integer sequence S₁,S₂,……,S_n. The score is the number of integers i satisfying 1 ≤ i ≤ n and S_i = D_i.

As a good tuner, DreamGrid can choose an integer (can be positive, 0, or negative) as his tune and add to every element in . Can you help him maximize his score by choosing a proper tune?
Input:
There are multiple test cases. The first line of the input contains an integer T (about 100), indicating the number of test cases. For each test case:

The first line contains one integer n (1 ≤ n ≤ 10⁵), indicating the length of the sequences D and S.

The second line contains n integers D₁,D₂,……,D_n (-10⁵ ≤ D_i ≤ 10⁵), indicating the song performed by DreamGrid.

The third line contains n integers S₁,S₂,……,S_n (-10⁵ ≤ S_i ≤ 10⁵), indicating the standard version of the song.

It’s guaranteed that at most 5 test cases have n > 100.
Output:
For each test case output one line containing one integer, indicating the maximum possible score.
Sample Input:
2
4
1 2 3 4
2 3 4 6
5
-5 -4 -3 -2 -1
5 4 3 2 1
Sample Output:
3

就是看两组数的差值想减,输出相同的数出现的最多次数,可能有负的,如果有负的可以转为正的.

看代码

# include <stdio.h>
# include <string.h>

int main(void)
{
	int t, n, a[100001], b[100001], c[300010];
	int i, j, min, m, sum, q, w, k;
	scanf("%d", &t);
	while (t --)
	{
		min = -1;
		memset(c, 0, sizeof(c)); //记得初始化
		scanf("%d", &n);
		for (i = 1; i <= n; i ++)
			scanf("%d", &a[i]);
		for (i = 1; i <= n; i ++)
		{
			scanf("%d", &b[i]);
			if (b[i]-a[i] < 0)
				c[-(b[i]-a[i]-100000)] ++;// 小于0的转化为1000001  比如-1为1000001 -2为10000002
			else 
			c[b[i]-a[i]] ++ ;
		}
		for (i = 0; i < 200100; i ++)
		{
			if (c[i] > min)
				min = c[i];//遍历一遍即可
		}
		
		printf("%d\n", min);
	}
	return 0;
}