Problem 1565 - B - Magic

最新推荐文章于 2025-08-04 14:19:44 发布

leibniz_zhang

最新推荐文章于 2025-08-04 14:19:44 发布

阅读量302

点赞数

CC 4.0 BY-SA版权

分类专栏：数学

本文链接：https://blog.youkuaiyun.com/leibniz_zhang/article/details/51103569

数学专栏收录该内容

59 篇文章

订阅专栏

本文探讨了一种策略，通过使用一次魔法操作来最大化数字序列的和。具体步骤包括选择一个区间，对区间内的每个元素应用特定公式，并最终计算调整后的最大总和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem 1565 - B - Magic
Time Limit: 1000MS Memory Limit: 65536KB
Total Submit: 477 Accepted: 180 Special Judge: No
Description
Here are n numbers.

You have a magic : first , you choose a interval [l,r],and then each Si(l<=i<=r) will be ( 10 – Si ) % 10.

You can use the magic at most once to make sum of all the numbers to be maximum.

What is the maximum sum you can get?

Input
First line of each case contains a number n.(1 ≤ n ≤ 1 000 000).

Next line contains n numbers without space-separated. Each position corresponds to a number of 0-9.
Output
Output the answer on a single line for each case.
Sample Input
10
3775053808
10
2580294019
10
4701956095
Sample Output
50
50
54
Hint

Source

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int shu[1001000];
int main()
{
	char ch[1001000];
	int n;
	while (~scanf("%d",&n))
	{
		scanf("%s",ch);
		int s,p;
		s=0;
		for (int i=0;i<n;i++)
		{
			p=ch[i]-'0';
			s+=p;
			shu[i]=(10-p)%10-p;
		}
		int ss=0,ka=0;
		for (int i=0;i<n;i++)
		{
			ka+=shu[i];
			ss=max(ss,ka);
			if (ka<0)
			ka=0; 
		}
	if (ss>0)
	s+=ss; 
		printf("%d\n",s);
	}
	return 0;
 }