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  • SGU107——987654321 problem

    For given number N you must output amount of N-digit numbers, such, that last digits of their square is equal to 987654321. 

     

    Input

    Input contains integer number N (1<=N<=106)

     

    Output

    Write answer to the output.

     

    Sample Input

    8
    

     

    Sample Output

    0

     

     

    简单数学,题目说,一个数字n、找n*n结果末尾是987654321的数。

     

    先傻瓜式搜索,发现8位以下没有符合要求的;9位有4个,10位以上根据数学排列知识就搞定了。

     

    最近在搞SGU,先从简单题目入手,碰到什么算法就搞什么算法。

     

     

    #include<iostream>
    #include<string.h>
    #include<stdio.h>
    #include<ctype.h>
    #include<algorithm>
    #include<stack>
    #include<queue>
    #include<set>
    #include<math.h>
    #include<vector>
    #include<map>
    #include<deque>
    #include<list>
    using namespace std;
    int main()
    {
    //    for(int i=0;i<99999999;i++)
    //    if(i*i==987654321)
    //    printf("%d
    ",i);
    //    for(long long i=sqrt(987654321.0);i<=999999999;++i)
    //        if( i*i%1000000000 == 987654321 )
    //            printf("%d
    ",i);
    //}
    //    111111111
    //    119357639
    //    380642361
    //    388888889
    //    611111111
    //    619357639
    //    880642361
    //    888888889
    
        int n;
        while (scanf("%d",&n)!=EOF)
        {
            if (n<=8)
                printf("0
    ");
            else if (n==9)
                printf("8
    ");
            else
            {
                printf("72");
                for (int i=0;i<n-10;++i)
                    printf("0");
                    printf("
    ");
            }
        }
    
    }
    


     

     

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  • 原文地址:https://www.cnblogs.com/fuhaots2009/p/3435844.html
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