Ignatius‘s puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6559 Accepted Submission(s): 4540
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
Author
eddy
题意：求出使x为任意值时f(x)都能被65整除的最小a的值，若不存在输出no。
题解：若任意f（x）都能被65整除，那么f(1)%65==0,(f(x+1)-f(x))%65 == 0,二项式展开得(18+k*a)%65 == 0,所以只需要遍历找到a即可。
#include <stdio.h>
int main() {
int k, a;
while(~scanf("%d", &k)) {
for(a = 0; a <= 65; ++a)
if((18 + k*a) % 65 == 0) {
printf("%d\n", a);
break;
}
if(a > 65) printf("no\n");
}
return 0;
}
HDU1098 Ignatius's puzzle 【数论】
原文：http://blog.csdn.net/chang_mu/article/details/40346883
