原文：
Description
People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that this market segment was so far underestimated and that there is lack of such games. This kind of game was thus included into the KOKODáKH. The rules follow: Each player chooses two numbers Ai and Bi and writes them on a slip of paper. Others cannot see the numbers. In a given moment all players show their numbers to the others. The goal is to determine the sum of all expressions AiBi
from all players including oneself and determine the remainder after division by a given number M. The winner is the one who first determines the correct result. According to the players' experience it is possible to increase the difficulty by choosing higher numbers. You should write a program that calculates the result and is able to find out who won the game.
Input
The input consists of Z assignments. The number of them is given by the single positive integer Z appearing on the first line of input. Then the assignements follow. Each assignement begins with line containing an integer M (1 <= M <= 45000). The sum will be divided by this number. Next line contains number of players H (1 <= H <= 45000). Next exactly H lines follow. On each line, there are exactly two numbers Ai and Bi separated by space. Both numbers cannot be equal zero at the same time.
Output
For each assingnement there is the only one line of output. On this line, there is a number, the result of expression (A1B1+A2B2+ ... +AHBH)mod M
.Sample Input
3
16
4
2 3
3 4
4 5
5 6
36123
1
2374859 3029382
17
1
3 18132
Sample Output
2
13195
13
提示：本题考察了快速幂的运算，同时也考察了公式(a+b) % p = (a%p+b%p) % p的运用。
代码：
#include<iostream> using namespace std; struct AB { int a; int b; }; struct AB ab[45000];
这道题因为a与b有对应关系，可以用结构体解决。
int powermod(int a,int b,int m) { int ans = 1; a = a % m; while (b > 0) { if(b % 2 != 0) { ans = (ans * a) % m; } b = b / 2; a = (a * a) % m; } return ans; }
快速幂的基本思想：反复平方。
以下是我个人的理解：
每一个十进制整数都可以用2的不同次方来表示，比如x的22次方可以表示为x的16次方与x的4次方与x的2次方的积。事实上，22的二进制表示为：10110，1代表2的某次方上的权值为1，0代表为0。因此，可以将b%2（每次提取b的最后一位（b以二进制数表示）），为1时记录结果，为0时跳过，后将a平方，将b的二进制的形式右移一位（b / 2）即可。
测试：
void test() { int i; int t; int m; int h;//lines; int ans; cin >> t; while(t--) { ans = 0; cin >> m; cin >> h; for(i = 0;i < h;i++) { cin >> ab[i].a >> ab[i].b; ans = ans + powermod(ab[i].a,ab[i].b,m); } ans = ans % m; cout << ans << endl; } } int main() { test(); return 0; }
注意：每一次都要求模，最后还要再对整个ans求模。
由于自身语言表达水平有限，请谅解。
如果有好的建议，欢迎交流。
