718AEfim and Strange Grade

最新推荐文章于 2019-06-11 16:29:29 发布

原创最新推荐文章于 2019-06-11 16:29:29 发布 · 395 阅读

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本文介绍了一个关于如何通过有限次数的操作来最大化一个正小数分数值的问题。主人公Efim尝试利用特定的四舍五入规则，在给定的时间内尽可能提高他的考试成绩。文章详细解释了具体的实现算法，并提供了一段C语言的代码示例。

Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any positive decimal fraction. First he got disappointed, as he expected a way more pleasant result. Then, he developed a tricky plan. Each second, he can ask his teacher to round the grade at any place after the decimal point (also, he can ask to round to the nearest integer).

There are t seconds left till the end of the break, so Efim has to act fast. Help him find what is the maximum grade he can get in no more than t seconds. Note, that he can choose to not use all t seconds. Moreover, he can even choose to not round the grade at all.

In this problem, classic rounding rules are used: while rounding number to the n-th digit one has to take a look at the digit n + 1. If it is less than 5 than the n-th digit remain unchanged while all subsequent digits are replaced with 0. Otherwise, if the n + 1 digit is greater or equal to 5, the digit at the position n is increased by 1 (this might also change some other digits, if this one was equal to 9) and all subsequent digits are replaced with 0. At the end, all trailing zeroes are thrown away.

For example, if the number 1.14 is rounded to the first decimal place, the result is 1.1, while if we round 1.5 to the nearest integer, the result is 2. Rounding number 1.299996121 in the fifth decimal place will result in number 1.3.

Input

The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 109) — the length of Efim's grade and the number of seconds till the end of the break respectively.

The second line contains the grade itself. It's guaranteed that the grade is a positive number, containing at least one digit after the decimal points, and it's representation doesn't finish with 0.

Output

Print the maximum grade that Efim can get in t seconds. Do not print trailing zeroes.

思路：从小数点开始往后找第一个大于5的点，然后从这个点在往前遍历，有一些特殊情况需要考虑，比如小数往整数部分进位，和小数后没有比5大的情况等等

#include<stdio.h>
#include<string.h>
char ch[200005];
int main()
{
	int t,dight,first,now=0,len,i,begin;
	scanf("%d %d",&len,&t);
	memset(ch,0,sizeof(ch));
	getchar();
	scanf("%s",ch+2);//数组前面留几个空是防止出现9.5这种情况 
	ch[0]=ch[1]='0';
	len+=2;
	begin=2;
	for(i=0;i<len;i++)
	if(ch[i]=='.')
	dight=i;
	first=dight;
	for(i=dight+1;i<len;i++)
	if(ch[i]>='5')
	{
		first=i;
		break;
	}
	if(t==0||first==dight);
	else
	{
		if(first==dight+1)
		{
			ch[dight-1]+=1;
			len=dight;
		}
		else
		{
			ch[first-1]+=1;
			len=first;
			now++;
			for(i=len-1;i>dight+1;i--)
			{
				if(now>=t)
				break;
				if(ch[i]>='5')
				{
					ch[i-1]+=1;
					len=i;
					now++;
				}	
			}
			if(now<t)
			{	
				if(ch[dight+1]>='5')
				{
					ch[dight-1]+=1;
					len=dight;
				}
			}
		}
		i=1;
		while(ch[dight-i]>'9')
			{
				ch[dight-i-1]+=1;
				ch[dight-i]-=10;	
				begin=(dight-i-1)<2?(dight-i-1):2;
				i++;
			}
	}
	for(i=begin;i<len;i++)
	printf("%c",ch[i]);
	printf("\n");
	return 0;
}