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社区首页 >专栏 >uva---(11549)CALCULATOR CONUNDRUM

uva---(11549)CALCULATOR CONUNDRUM

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Gxjun
发布2018-03-26 10:25:19
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发布2018-03-26 10:25:19
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文章被收录于专栏:ml

Problem C

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

代码语言:javascript
复制
2
1 6
2 99

OUTPUT

代码语言:javascript
复制
9
99

模拟题:
 题意就是一个数,然后一直平方,然后求在这个过程中最大的前n位:
代码:
代码语言:javascript
复制
 1 #include<cstring>
 2 #include<set>
 3 #include<iostream>
 4 using namespace std;
 5 long long mod[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000};
 6 int main(){
 7     int n,test;
 8     long long k;
 9     cin>>test;
10     while(test--){
11       cin>>n>>k;
12       while(k/mod[n]>0){  //截取高位
13            k/=10;
14        }
15      long long res=k;
16      set<long long>sac;
17       while(sac.count(k)==0){
18         sac.insert(k);
19          k*=k;
20         while(k/mod[n]>0){  //截取高位
21            k/=10;
22        }
23         if(k>res)res=k;
24      }
25      cout<<res<<endl;
26     }
27   return 0;
28 }
代码语言:javascript
复制
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原始发表:2014-08-18 ,如有侵权请联系 cloudcommunity@tencent.com 删除

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