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HDU_1098 Ignatius's puzzle[规律题]
Ignatius‘s puzzle
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".
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
题意:f(x)=5*x^13+13*x^5+k*a*x,找出最小的a,使得对于给定的k,任意x下f(x)能被65整除。
思路:打表找规律。
你会发现,(5*x^13+13*x^5)%65每65个x成循环。要使结果成立(k*a*x)%65,也得每65个循环,且a的值在0-64之间,所以就有如下代码-> _ ->
代码:
/************************************************ * Author: Ac_sorry * File: * Create Date: * Motto: One heart One life * CSDN: http://blog.csdn.net/code_or_code *************************************************/ #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<cstdlib> #include<vector> #include<string> #include<map> #include<utility> #include<queue> #include<stack> #define INF 0x3f3f3f3f #define MOD 1000000007 #define seed_ 131 #define eps 1e-8 #define mem(a,b) memset(a,b,sizeof a) #define w(i) T[i].w #define ls(i) T[i].ls #define rs(i) T[i].rs using namespace std; typedef long long LL; const int N=100010; int a[100]; int main() { //int T;scanf("%d",&T); //int AC=0; for(int x=1;x<=65;x++) a[x]=(5*x%65*x*x%65*x*x%65*x*x%65*x*x%65*x*x%65*x*x%65+13*x%65*x*x%65*x*x%65)%65; int k; while(scanf("%d",&k)==1) { int ok,ans; for(int aa=0;aa<65;aa++) { ok=1; for(int i=1;i<=65;i++) { if((a[i]+k*i%65*aa%65)%65!=0) { ok=0; break; } } if(ok) { ans=aa; break; } } if(ok) printf("%d\n",ans); else printf("no\n"); } return 0; }
HDU_1098 Ignatius's puzzle[规律题]
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