杭電ACM——1098

Ignatius's puzzle

問題： 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".

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

輸出：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.

示例輸入： 11 100 9999

示例輸出： 22 no 43

問題分析：            

5*x^13+13*x^5+k*a*x可以分成兩部分 即 5*x^13+13*x^5 和 k*a*x;      

而第一個部分我們可以發現 （5*x^13+13*x^5）% 65 結果會不斷循環，循環節為65，而且為等差數列，差為18，則通過數學歸納法證明可得      

隻要滿足（18 + k * x） % 65 == 0即可      

代碼如下      

# include <stdio.h>


int main()
{
	int k, flag, x;
	while(scanf("%d", &k) != EOF)
	{
		flag = 0;
		for(x = 1; x <= 65; x++)
		{
			if ((18 + k * x) % 65 == 0) 
			{
				flag = 1;
				printf("%d\n", x);
				break;
			}
		}
		if (flag == 0)
		{
			printf("no\n");
		}
	}


	return 0;
}