zoukankan      html  css  js  c++  java
  • uva---(11549)CALCULATOR CONUNDRUM

    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

    2
    1 6
    2 99
    

    OUTPUT

    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:
     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 }
    View Code
     
  • 相关阅读:
    day22【网络编程】
    day21【缓冲流、转换流、序列化流】
    day20【字节流、字符流】
    设计模式7-适配器模式
    设计模式6-状态模式
    设计模式5-观察者模式
    设计模式4-建造者模式
    Web Service与WCF与Web API区别
    设计模式3-外观模式
    设计模式2-模板方法模式
  • 原文地址:https://www.cnblogs.com/gongxijun/p/3918952.html
Copyright © 2011-2022 走看看