zoukankan      html  css  js  c++  java
  • 【hdu 2662 Coin】

    Coin

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 662    Accepted Submission(s): 329


    Problem Description
    Moon has many coins, but only contains two value types which is 5 cents and 7 cents, Some day he find that he can get any value which greater than 23 cents using some of his coins. For instance, he can get 24 cents using two 5 cents coins and two 7 cents coins, he can get 25 cents using five 5 cents coins, he can get 26 cents using one 5 cents coins and three 7 cents coins and so on. 

    Now, give you many coins which just contains two value types just like Moon, and the two value types identified by two different prime number i and j. Can you caculate the integer n that any value greater than n can be created by some of the given coins.
     
    Input
    The first line contains an integer T, indicates the number of test cases. 
    For each test case, there are two different prime i and j separated by a single space.(2<=i<=1000000, 2<=j<=1000000)
     
    Output
    For each test case, output one line contains the number n adapt the problem description.
     
    Sample Input
    1 5 7
     
    Sample Output
    23
     
    Source
     
    Recommend
    zty
     
     
     
     1 #include <iostream>
     2 using namespace std;
     3 
     4 typedef unsigned long long int longint;
     5 
     6 
     7 int main()
     8 {
     9     int t;
    10     cin >> t;
    11     while (t--)
    12     {
    13         longint n, m;
    14         cin >> n >> m;
    15         longint ans;
    16 
    17         ans = (n * m) - n - m;
    18 
    19         cout << ans << endl;
    20     }
    21     return 0;
    22 }
    23 
    24 // end
    25 // ism

    出几组数据就可以猜到a*b-a-b,不过这里还是证明一下吧。

    设所求为n,那么n+a、n+b可以用a、b线性表出,而n不可。
    所以 n+a=x1*a+y1*b,n+b=x2*a+y2*b
    所以 n=(x1-1)*a+y1*b n=x2*a+(y2-1)*b
    因为n不能被线性表出,所以x1=0,y2=0
    所以 n+a=y1*b,n+b=x2*a
    所以 n+a=y1*b,n+a=(x2+1)*a-b
    所以 (x2+1)*a-b是b的倍数
    因为a、b互质,所以(x2+1)是b的倍数
    因为求最小的n,所以选最小的x2值,所以取(x2+1)为b
    所以 n+a=b*a-b,n=a*b-a-b
    证毕

     

     

  • 相关阅读:
    finder的隐藏文件&IOS虚拟机地址
    IOS的UI总结
    ios系统的中arm指令集
    mac下删除svn账号
    PNG图片压缩工具
    让finder显示路径
    Serilog高级玩法之用Serilog记录所选终结点附加属性
    如何利用Serilog的RequestLogging来精简ASP.NET Core的日志输出
    关于C#异步编程你应该了解的几点建议
    C#异步编程入门看这篇就够了
  • 原文地址:https://www.cnblogs.com/ismdeep/p/2600867.html
Copyright © 2011-2022 走看看