HDU 1098 HDOJ 1098 Ignatius's puzzle ACM 1098 IN HDU

 

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题目地址:

     http://acm.hdu.edu.cn/showproblem.php?pid=1098

题目分析:

     纯粹的数学题, 数学归纳法的 应用  , 最后 归纳得出  原题等价与 18 + k*a 是否能被65整除.

代码如下 :

/*
Mail to   : miyubai@gamil.com
My Blog   : www.baiyun.me
Link      : http://www.cnblogs.com/MiYu  || http://www.cppblog.com/MiYu
Author By : MiYu
Test      : 1
Complier  : g++ mingw32-3.4.2
Program   : HDU_1098
Doc Name  : Ignatius's puzzle
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
inline bool scan_d(int &num)
{
        char in;bool IsN=false;
        in=getchar();
        if(in==EOF) return false;
        while(in!='-'&&(in<'0'||in>'9')) in=getchar();
        if(in=='-'){ IsN=true;num=0;}
        else num=in-'0';
        while(in=getchar(),in>='0'&&in<='9'){
                num*=10,num+=in-'0';
        }
        if(IsN) num=-num;
        return true;
}
int main ()
{
    int N;
    while ( scan_d ( N ) ) {
        if ( N % 65 == 0 )
            puts ( "no" );
        else {
            int i;
            for ( i = 0; i < 65; ++ i ) {
                if ( ( i * N + 18 ) % 65 == 0 ) {
                    printf ( "%d\n", i );
                    break;
                }
            }
            if ( i == 65 ) puts ( "no" );
        }
    }
    return 0;
}