HDU 2035

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”

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

这个题目注意两点：

(1)取a的后三位计算即可： a%1000

(2)会使用取模定理： (a*b)%1000 == ((a%1000) * (b%1000)) % 1000

如果，不考虑（2）的话，会溢出（long long都会溢出）.

#include <iostream>

using namespace std;

int main () {
int a, b, i, t;
while(cin >> a >> b, a|b) {
t = 1;
for(int i = 0; i < b; i++) {
t = (a%1000)*(t%1000)%1000;
}
cout << t << endl;
}
return 0;
}

下面这种是根据快速幂运算的算法进行的修改。

因为当A和B非常大时，就超出int的范围了。所以这题考的是数学技巧。首先，45896的7次方的后三位和896的7次方的后三位是一样的。所以对于输入的A进行%1000的处理。

#include <iostream>
using namespacestd;
intplay(intn,intm)
{
<wbr><wbr><wbr>if</wbr></wbr></wbr>(m==1)
<wbr><wbr><wbr>return</wbr></wbr></wbr>(n%1000);
<wbr><wbr><wbr>else if</wbr></wbr></wbr>(m%2==0)
<wbr><wbr><wbr>{</wbr></wbr></wbr>
<wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr>int</wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr>x;
<wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr>x<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">=</span></strong>play<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">(</span></strong>n<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">,</span></strong>m<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">/</span></strong><span style="color:#CC3300; word-wrap:normal; word-break:normal; line-height:24px">2</span><strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">);</span></strong><br><wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr>x<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">%=</span></strong><span style="color:#CC3300; word-wrap:normal; word-break:normal; line-height:24px">1000</span><strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">;</span></strong></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr>
<wbr><wbr><wbr><wbr><wbr><wbr><wbr><wbr>return</wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr>((x*x)%1000);
<wbr><wbr><wbr>}</wbr></wbr></wbr>
<wbr><wbr><wbr>else</wbr></wbr></wbr>
<wbr><wbr><wbr>{</wbr></wbr></wbr>
<wbr><wbr><wbr><wbr><wbr><wbr><wbr>int</wbr></wbr></wbr></wbr></wbr></wbr></wbr>s;<wbr><br><wbr><wbr><wbr><wbr><wbr><wbr><wbr>s<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">=</span></strong>play<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">(</span></strong>n<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">,</span></strong>m<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">/</span></strong><span style="color:#CC3300; word-wrap:normal; word-break:normal; line-height:24px">2</span><strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">);</span></strong><br><wbr><wbr><wbr><wbr><wbr><wbr><wbr>s<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">%=</span></strong><span style="color:#CC3300; word-wrap:normal; word-break:normal; line-height:24px">1000</span><strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">;</span></strong></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr>
<wbr><wbr><wbr><wbr><wbr><wbr><wbr>return</wbr></wbr></wbr></wbr></wbr></wbr></wbr>((n*s*s)%1000);
<wbr><wbr><wbr>}<br> }</wbr></wbr></wbr>
intmain()
{
<wbr><wbr><wbr>int</wbr></wbr></wbr>a,b;
<wbr><wbr><wbr>while</wbr></wbr></wbr>(cin>>a>>b)
<wbr><wbr><wbr>{</wbr></wbr></wbr>
<wbr><wbr><wbr><wbr><wbr><wbr>if</wbr></wbr></wbr></wbr></wbr></wbr>(a==0&&b==0)
<wbr><wbr><wbr><wbr><wbr><wbr>break</wbr></wbr></wbr></wbr></wbr></wbr>;
<wbr><wbr><wbr><wbr><wbr><wbr>int</wbr></wbr></wbr></wbr></wbr></wbr>sum(0);
<wbr><wbr><wbr><wbr><wbr><wbr>a<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">%=</span></strong><span style="color:#CC3300; word-wrap:normal; word-break:normal; line-height:24px">1000</span><strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">;</span></strong><br><wbr><wbr><wbr><wbr><wbr><wbr>sum<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">=</span></strong>play<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">(</span></strong>a<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">,</span></strong>b<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">);</span></strong><br><wbr><wbr><wbr><wbr><wbr><wbr>cout<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">&lt;&lt;</span></strong>sum<strong><span style="color:#FF00FF; word-wrap:normal; word-break:normal; line-height:24px">&lt;&lt;</span></strong>endl</wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr></wbr>;
<wbr><wbr><wbr>}</wbr></wbr></wbr>
<wbr><wbr><wbr>return</wbr></wbr></wbr>0;
}
什么是对快速求幂运算算法的一个修改