HDU 1395 2^x mod n = 1(暴力枚举)
2016-05-04 19:49
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2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 15721 Accepted Submission(s): 4870
[align=left]Problem Description[/align]
Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.
[align=left]Input[/align]
One positive integer on each line, the value of n.
[align=left]Output[/align]
If the minimum x exists, print a line with 2^x mod n = 1.
Print 2^? mod n = 1 otherwise.
You should replace x and n with specific numbers.
[align=left]Sample Input[/align]
2 5
[align=left]Sample Output[/align]
2^? mod 2 = 1 2^4 mod 5 = 1
[align=left]Author[/align]
MA, Xiao
[align=left]Source[/align]
ZOJ Monthly, February 2003
题解:暴力枚举....
TLE代码:
#include<iostream> #include<cstdlib> #include<cstdio> #include<cmath> #include<cstring> #include<string> #include<cstdlib> #include<iomanip> #include<algorithm> #include<time.h> typedef long long LL; using namespace std; int main() { int x,n; while(cin>>n) { int k=0; for(int i=1;i<=5000;i++) { if((int)pow(2,(double)i)%n==1){ printf("2^%d mod %d = 1\n",i,n);break; } k++; } if(k==5000) cout<<"2^? mod "<<n<<" = 1"<<endl; } return 0; }
AC代码:
#include<iostream> #include<cstdlib> #include<cstdio> #include<cmath> #include<cstring> #include<string> #include<cstdlib> #include<iomanip> #include<algorithm> #include<time.h> typedef long long LL; using namespace std; int main() { int k,m,x,n; while(cin>>n) { k=1; x=2; if(n%2==0||n==1) cout<<"2^? mod "<<n<<" = 1"<<endl; else { while(k++) { x*=2; x=x%n; if(x==1) break; } cout<<"2^"<<k<<" mod "<<n<<" = 1"<<endl; } } return 0; }
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