人见人爱A^B
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35507 Accepted Submission(s): 24132
Problem Description
求A^B的最后三位数表示的整数。
说明:A^B的含义是“A的B次方”
说明:A^B的含义是“A的B次方”
Input
输入数据包含多个测试实例,每个实例占一行,由两个正整数A和B组成(1<=A,B<=10000),如果A=0, B=0,则表示输入数据的结束,不做处理。
Output
对于每个测试实例,请输出A^B的最后三位表示的整数,每个输出占一行。
Sample Input
2 3 12 6 6789 10000 0 0
Sample Output
8 984 1
二分法快速乘方
#include<stdio.h>
#define LL long long
LL quickPower(LL a,LL b)
{
LL ans=1;
if(b==0)
return 1;
if(b<0)
return 0;
while(b)
{
if(b&1)
ans*=a;
a*=a;
b>>=1;
}
return ans;
}
int main()
{
LL n,m;
//freopen("in.txt","r",stdin);
while(scanf("%I64d%I64d",&n,&m)!=EOF)
{
printf("%I64d\n",quickPower(n,m));
}
return 0;
}
最简单粗暴的方法:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
//freopen("in.txt","r",stdin);
int n,m;
while(scanf("%d%d",&n,&m),n+m!=0)
{
int ans=1;
for(int i=0;i<m;i++){
ans*=n;
ans%=1000;
}
printf("%d\n",ans);
}
return 0;
}
二分法优化:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define LL long long
LL mod(LL a,LL n,LL m)
{
LL temp=a,ans=1;
while(n)
{
if(n&1)
ans=ans*temp%m;
temp=temp*temp%m;
n>>=1;
}
return ans;
}
int main()
{
//freopen("in.txt","r",stdin);
int n,m;
while(scanf("%d%d",&n,&m),n+m!=0)
{
printf("%lld\n",mod(n,m,1000));
}
return 0;
}