poj1995快速幂

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

题目大意

求a^b %p
1≤a,b,p≤10^9

思路

>

时间O(10^9)一定会爆T，采用数学方法+位运算，得到O（log b)的快速幂算法

代码

#include<cstdio>
#include<iostream>
#include<cctype>
#include<algorithm>
#define ll long long
using namespace std;

inline int read()
{
    int ans=0,f=1;
    char chr=getchar();
    while(!isdigit(chr)) {if(chr='-') f=-1; chr=getchar();}
    while(isdigit(chr)) {ans=ans*10+chr-'0'; chr=getchar();}
    return ans*f;
}
ll T,p; 
int n;

ll calc(ll a,ll b,ll c)//计算a^b%p
{
    ll tans=1;//记录答案
    while(b)
    {
        if(b&1) tans=tans*a%p;//如果是奇数，意味着这一处数位要取，更新ans;
        a=a*a%p;//更新a；
        b>>=1;//将b左移一位
    }
    return tans;
}
int main()
{
    T=read();
    while(T--)
    {
        ll ans=0;
        p=read();
        n=read();
        for(int i=1;i<=n;i++)
        {
            ll a=read(),b=read();
            ans=(ans+calc(a,b,p))%p;
        }
        printf("%d\n",ans);
    }
    return 0;
}