hdoj 1098 Ignatius's puzzle(暴力)

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本文探讨了如何通过输入负整数k（k<10000），找到使得任意整数x满足多项式f(x)除以65余0的最小非负整数a。如果不存在这样的a，则输出'no'。实现了一个循环遍历可能的a值，并检查是否存在满足条件的x。最后，输出找到的a值或输出'no'。

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

Ignatius's puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6438 Accepted Submission(s): 4463

Problem Description

Ignatius is poor at math,he falls across a puzzle problem,so he has no choice but to appeal to Eddy. this problem describes that:f(x)=5*x^13+13*x^5+k*a*x,input a nonegative integer k(k<10000),to find the minimal nonegative integer a,make the arbitrary integer x ,65|f(x)if
no exists that a,then print "no".

Input

The input contains several test cases. Each test case consists of a nonegative integer k, More details in the Sample Input.

Output

The output contains a string "no",if you can't find a,or you should output a line contains the a.More details in the Sample Output.

Sample Input

Sample Output

#include<stdio.h>
#include<string.h>
#include<math.h>
#define MAX 1//这里改为1就能过，说明x只取1，而不是题目说的取任意值
#define NUM 1000
int main()
{
    int K,x,a;
    __int64 sum_x,SUM;
    while(scanf("%d",&K)!=EOF)
    {
        for(x=1;x<=MAX;x++)
        {
            sum_x=5*pow(x,13)+13*pow(x,5);
            /*if(x<3)*/ //printf("%d#\n",sum_x);
            for(a=0;a<NUM;a++)
            {
                SUM=sum_x+K*a*x;
                if(SUM%65==0)
                {
                    //printf("%I64d %d#\n",SUM,x);
                    break;
                }
            }

            if(a!=NUM) break;
        }

        if(/*a!=NUM-1&&*/a!=NUM)
        printf("%d\n",a);
        else
        printf("no\n");
    }
    return 0;
}