uva11549 Calculator Conundrum【解法一】

最新推荐文章于 2017-11-23 13:41:54 发布

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最新推荐文章于 2017-11-23 13:41:54 发布

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分类专栏：其他算法 UVa 文章标签：模拟哈希

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本文介绍了一个基于旧计算器的数字游戏，玩家通过不断平方初始数字并截取最显著位来获得最大可能数值。文中提供了一段C++代码实现，利用了模运算和哈希表来避免重复计算，最终输出在给定显示位数限制下能达到的最大数字。

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 over ows. When
the result over ows, only the n most signi cant digits are displayed
on the screen and an error ag 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 rst 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< 10 n
) 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.

一个一个模拟，然后哈希判重。注意多用几个质数取模。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
int p[]={1000007,133331,324617,287933,77447,145009,151939,152083,159113,159119},
a[20],n;
bool v[15][1000010];
int ne(int x)
{
    int l=0,ret=0,i;
    LL t=(LL)x*x;
    a[1]=0;
    while (t)
    {
        a[++l]=t%10;
        t/=10;
    }
    for (i=l;i&&i>=l-n+1;i--)
      ret=ret*10+a[i];
    return ret;
}
int main()
{
    int k,ans,T,i;
    bool flag;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&n,&k);
        memset(v,0,sizeof(v));
        ans=0;
        while (1)
        {
            flag=1;
            for (i=0;i<10;i++)
              if (!v[i][k%p[i]])
              {
                flag=0;
                break;
              }
            if (flag) break;
            for (i=0;i<10;i++)
              v[i][k%p[i]]=1;
            ans=max(ans,k);
            k=ne(k);
        }
        printf("%d\n",ans);
    }
}