Ignatius's puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8929    Accepted Submission(s): 6204 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 11 100 9999   Sample Output 22 no 43

用数学归纳法做。

f(x) = 5*x^13 + 13*x^5 + k*a*x

f(x + 1) - f(x) = 5*x^13 + 13*x^5 + k*a*x + 65*(x^12 + x^11 + ..... + x) + 65*(x^5 + x^4 + ..... + x) + 18 + k*a.

所以是否能整除65由18 + k*a决定。

#include <iostream> #include <cstdio> #include <cstring> using namespace std; int main() { int k; while (~scanf("%d", &k)) { int ans = -1; for (int i = 0; i < 66; ++i) { if ((k * i + 18) % 65 == 0) { ans = i; break; } } if (ans != -1) printf("%d\n", ans); else printf("no\n"); } return 0; } 有小伙伴不明白为什么为什么0 <= a < 66

嗯。

(18 + k*a) % 65 = (18 % 65) + (k * a) % 65 = 18 + (k % 65) * (a % 65)

所以、a超过65是没有意义的