UVA 11549 Calculator Conundrum

题意:有个老式计算器,每次只能记住一个数字的前n位。现在输入一个整数k,然后反复平方,一直做下去,能得到的最大数是多少。例如,n=1,k=6,那么一次显示:6,3,9,1...

思路:直接开个set判重即可


#include<cstdio>
#include<set>
using namespace std;
int T,n,k;
int next(int x)//寻找x的后继
{
	if(!k) return 0;
	int buf[20],i=0;
	long long t=(long long)x*x;
	while(t)
	{
		buf[i++]=t%10;
		t/=10;
	}
	int ans=0;
	for(int j=1;j<=min(n,i);++j) ans=ans*10+buf[i-j];
	return ans;
}
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&k);
		set<int> S;
		while(!S.count(k))
		{
			S.insert(k);
			k=next(k);
		}
		printf("%d\n",*(--S.end()));
	}
	return 0;
}

Alice got a hold of an old calculator that can display n digits. She was bored enough to come up with
the following time waster.
She enters a number k then repeatedly squares it until the result overflows. When the result
overflows, only the n most significant digits are displayed on the screen and an error flag appears. Alice
can clear the error and continue squaring the displayed number. She got bored by this soon enough,
but wondered:
“Given n and k, what is the largest number I can get by wasting time in this manner?”
Input
The first line of the input contains an integer t (1 ≤ t ≤ 200), the number of test cases. Each test case
contains two integers n (1 ≤ n ≤ 9) and k (0 ≤ k < 10n) where n is the number of digits this calculator
can display k is the starting number.
Output
For each test case, print the maximum number that Alice can get by repeatedly squaring the starting
number as described.
Sample Input
2
1 6
2 99
Sample Output
9
99

posted @ 2016-04-06 09:15  围巾的ACM  阅读(147)  评论(0编辑  收藏  举报