HDU 1098 （数学_特殊值+抽屉原理）

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题目大意：方程f(x)=5*x^13+13*x^5+k*a*x；输入任意一个数k，是否存在一个数a，对任意x都能使得f(x)能被65整出。现假设存在这个数a ,因为对于任意x方程都成立所以，当x=1时f(x)=18+ka又因为f(x)能被65整出，故设n为整数可得，f(x)=n*65;即：18+ka=n*65;因为n为整数，若要方程成立则问题转化为，对于给定范围的a只需要验证，是否存在一个a使得(18+k*a)%65==0所以容易解得注意，这里有童鞋不理解为毛a只需到65即可因为，当a==66时也就相当于已经找了一个周期了，所以再找下去也找不到适当的a了如果你非要证明的话，可以利用了取模过程与数的运算的次序上可交换原理简单证明一下
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)ifno 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
22no43
 


#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;int main(){int a,k;while(scanf("%d",&k)!=EOF) {for(a=1;a<=65;a++) {if((18+a*k)%65==0) break;}if(a>65) printf("no\n");else printf("%d\n",a);}return 0;}
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