HDU1098 Ignatius's puzzle

1098

Ignatius’s puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9936 Accepted Submission(s): 6961

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

Author
eddy

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f(x)=5*x^13+13*x^5+k*a*x，要使f(x)被65整除f（x）肯定为整数，—-》x为整数！=0，
f(x+1)=5*(x+1)^13+13*(x+1)^5+k*a*(x+1),将f(x+1)按二项式定理展开有：
f(x+1)=5*(c(13,0)*x^13+c(13,1)*x^12+c(13,2)*x^11+….+c(13,12)*x+c(13,13)*x^0)+
13*(c(5,0)x^5+c(5,1)*x^4+…..+c(5,4)*x^1+c(5,5)*x^0)+k*a(x+1)
由于c(13,1)…c(13,12)中间一定可以提取一个13,则有这些项*5之后一定可以被65整除
同理c(5,1)…c(5,4)一定可以提取一个5,则有这些项*13之后一定可以被65整除
所以：f(x+1)=5*(c(13,0)x^13+c(13,13)*x^0)+13(c(5,0)x^5+c(5,5)*x^0)+k*a(x+1)
只需要k*a*（x+1）能被65整除，即k*a*x能被65整除，要想取a最小值，x要取最小1或者-1。所以只需要18+k*a或者-18-k*a能被65整除。要使(18+k*a)%65==0，k*a肯定为65的倍数-18=47，而k最小为1.所以a最大为65 就可以了。
原文

#include<stdio.h>#include<math.h>using namespace std;int main() {    int k;    while (scanf("%d",&k) != EOF)    {        bool flag = true;        for (int i = 1; i <= 65; i++)        {            if (i*k % 65 == 47)            {                printf("%d\n", i);                flag = false;                break;            }        }        if (flag)printf("no\n");    }    return 0;}