UVA - 11549 Calculator Conundrum-优快云博客

CALCULATOR CONUNDRUM

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?”

Program 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.

Program Output

For each test case, print the maximum number that Alice can get by repeatedly squaring the starting number as described.

Sample Input & Output

INPUT

 2 1 6 2 99

OUTPUT

9
99

Calgary Collegiate Programming Contest 2008

分析：

刘汝佳大白书里的简单题。需要注意的是：k的范围太大，用数组开不下，所以用STL的集合。

解法一：

耗时4.5s（本题要求6s，所以这种方法也可以ac）

ac代码：

#include <iostream>
#include<set>
#include<cstdio>
#include<sstream>
using namespace std;

int next(int n,int k)
{
stringstream ss;
ss<<(long long)k*k;
string s=ss.str();
if(s.length()>n) s=s.substr(0,n);
int ans;
stringstream ss2(s);//
ss2>>ans;
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,k;
scanf("%d%d",&n,&k);
set<int>s;
int ans=k;
while(!s.count(k))
{
s.insert(k);
if(k>ans) ans=k;
k=next(n,k);
}
printf("%d\n",ans);
}
return 0;
}

解法二：

STL的string很慢，stringstream更慢，所以考虑把它们换掉，时间优化到1s

ac代码：

#include <iostream>
#include<set>
#include<cstdio>
#include<sstream>
using namespace std;
const int maxn=100;
int buf[maxn];
int next(int n,int k)
{
//if(!k) return 0;//这句可以省
long long k2=(long long)k*k;
int L=0;
while(k2>0)
{
buf[L++]=k2%10;
k2/=10;
}
if(n>L) n=L;
int ans=0;
for(int i=0;i<n;i++)
ans=ans*10+buf[--L];
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,k;
scanf("%d%d",&n,&k);
set<int>s;
int ans=k;
while(!s.count(k))
{
s.insert(k);
if(k>ans) ans=k;
k=next(n,k);
}
printf("%d\n",ans);
}
return 0;
}

解法三：

STL的set还是慢，考虑把set也换掉，用Floyd判圈算法，时间优化到0.5s，空间复杂度降到O（1）.

这种方法初学还是感到很巧妙的。

ac代码：

#include <iostream>
#include<cstdio>
using namespace std;
const int maxn=100;
int buf[maxn];

int next(int n,int k)
{
//if(!k) return 0;//
long long k2=(long long)k*k;
int L=0;
while(k2>0)
{
buf[L++]=k2%10;
k2/=10;
}
if(n>L) n=L;
int ans=0;
for(int i=0;i<n;i++)
ans=ans*10+buf[--L];
return ans;
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,k;
scanf("%d%d",&n,&k);
int ans=k;
int k1=k,k2=k;
do
{
k1=next(n,k1);
k2=next(n,k2);
if(k2>ans)
ans=k2;
k2=next(n,k2);
if(k2>ans)
ans=k2;
}while(k1!=k2);
printf("%d\n",ans);
}
return 0;
}