HDOJ 题目4394 Digital Square（DFS）

最新推荐文章于 2019-08-02 23:48:30 发布

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本文介绍了一道关于寻找最小非负整数M的问题，该整数的平方尾数匹配给定整数N的任意位数。文章通过提供详细的算法实现代码，包括初始化、解析输入数字和深度优先搜索策略，展示了如何有效地解决这个问题。

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Digital Square

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1757 Accepted Submission(s): 677

Problem Description

Given an integer N,you should come up with the minimum nonnegative integer M.M meets the follow condition: M ²%10 ^x=N (x=0,1,2,3....)

Input

The first line has an integer T( T< = 1000), the number of test cases.
For each case, each line contains one integer N(0<= N <=10 ⁹), indicating the given number.

Output

For each case output the answer if it exists, otherwise print “None”.

Sample Input

Sample Output

  
   None
11
5

Source

2012 Multi-University Training Contest 10

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ac代码

#include<stdio.h>
#include<string.h>
#define INF 1<<30
#define max(a,b) (a>b?a:b)
#define min(a,b) (a>b?b:a)
int a[20];
__int64 mod[20],ans,n,len;
void init()
{
	mod[0]=1;
	for(int i=1;i<20;i++)
	{
		mod[i]=mod[i-1]*10;
	}
}
void fun()
{
	__int64 temp=n;
	len=0;
	memset(a,0,sizeof(a));
	while(temp)
	{
		a[len++]=temp%10;
		temp/=10;
	}
}
void dfs(int now,int step)
{
	if(step==len)
	{
		ans=min(ans,now);
		return;
	}
	for(int i=0;i<=9;i++)
	{
		__int64 to,temp=now+mod[step]*i;
		to=temp;
		temp*=temp;
		temp/=mod[step];
		temp%=10;
		if(temp==a[step])
		{
			dfs(to,step+1);
		}
	}
}
int main()
{
	int t;
	init();
	scanf("%d",&t);
	while(t--)
	{
		scanf("%I64d",&n);
		fun();
		ans=INF;
		dfs(0,0);
		if(ans==INF)
		{
			printf("None\n");
		}
		else
			printf("%d\n",ans);
	}
}