杭电acm1098 Ignatius's puzzle

最新推荐文章于 2020-04-11 15:33:45 发布

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最新推荐文章于 2020-04-11 15:33:45 发布

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杭电acm1098

Ignatius's puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3894 Accepted Submission(s): 2608

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

111009999

Sample Output

Author

eddy

题目中文意思：

一个函数：  f(x)=5*x^13+13*x^5+k*a*x
给定一个非负的k值 ，求最小的非负的a值，(k, a 都是整型)使得对任意的整数x都能使f(x)被65整除。
每输入一个k 值 ，对应输出一个 a值 ，若a值不存在的话，则输出 no

解题思路：数学归纳法 f(x)=5*x^13+13*x^5+k*a*x
假设当x=n时 65|f(x)成立，那一定有 65|f(n+1)成立
那么65|f(n+1)-f(n) 成立通过二项展开可以得到只要18+k*a可以被65整除就可以了
在1~65之间遍历就ok。

#include <stdio.h>
int main()
{
int a,k,sum;
while(scanf("%d",&k)!=EOF)
{
    for(a=1;a<=65;a++)
    {
       sum = 18 + a*k;
    if(sum%65 == 0)
     break;
    }
    if(a == 66)
        printf("no\n");
       else
     printf("%d\n",a);
}
return 0;
}