首页 > 代码库 > 快速幂 生成素数表
快速幂 生成素数表
A hard puzzle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 30672 Accepted Submission(s): 11006
Problem Description
lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
this puzzle describes that: gave a and b,how to know the a^b‘s the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.
this puzzle describes that: gave a and b,how to know the a^b‘s the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.
Input
There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)
Output
For each test case, you should output the a^b‘s last digit number.
Sample Input
7 66 8 800
Sample Output
9 6
#include<iostream>
#include<cstdio>
#include<cmath>
#define MOD 10
#define LL long long
using namespace std;
int pow_mod(LL a,LL b)
{
int num=1;
while(b)
{
if(b&1) num=(num*a)%MOD;
a=(a*a)%MOD;
b>>=1;
}
return num;
}
int main()
{
int a,b;
while(scanf("%d%d",&a,&b)!=EOF)
{
printf("%d\n",pow_mod(a,b));
}
return 0;
}
生成素数表：
void Prime(int x)
{
for(int i=2;i<=x;i++)
{
if(flag[i]) continue;
prime[cnt++]=i;
for(int j=2;i*j<=x;j++)
flag[i*j]=1;
}
}快速幂 生成素数表
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。