HDU 5666 Segment（BestCoder Round #80 1002）

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Segment

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 787    Accepted Submission(s): 301

Problem Description

     Silen August does not like to talk with others.She like to find some interesting problems.

     Today she finds an interesting problem.She finds a segment  x+y=q .The segment intersect the axis and produce a delta.She links some line between  (0,0)  and the node on the segment whose coordinate are integers.

     Please calculate how many nodes are in the delta and not on the segments,output answer mod P.

 

Input

     First line has a number,T,means testcase number.

     Then,each line has two integers q,P.

    q  is a prime number,and  2≤q≤1018,1≤P≤1018,1≤T≤10.

 

Output

     Output 1 number to each testcase,answer mod P.

 

Sample Input

   
   
   
   

    
    
    
    1 2 107
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0
   
   
   
   

 

Source

BestCoder Round #80

题目大意：

    Rivendell非常神,喜欢研究奇怪的问题.

\ \ \ \    今天他发现了一个有趣的问题.找到一条线段$x+y=q$,令它和坐标轴在第一象限围成了一个三角形,然后画线连接了坐标原点和线段上坐标为整数的格点.

\ \ \ \    请你找一找有多少点在三角形的内部且不是线段上的点,并将这个个数对$P$取模后告诉他.

官方题解：

考虑一条以$(0,0)$为起点,$(x,y)$为终点的线段上格点的个数（不包含端点时）,一定是$gcd(x,y)-1$,这个很显然吧.

然后整个网格图范围内的格点数目是\frac {q*(q-1)} 2​2​​q∗(q−1)​​.

所以答案就是\frac {q*(q-1)} 2 -​2​​q∗(q−1)​​− 所有线段上的格点的个数.

因为$gcd(a,b)=gcd(a,b-a)(b>a)$,所以$gcd(x,y)=gcd(x,p-x)=gcd(x,p)$,p是质数,所以$gcd(x,y)=1$,所以线段上都没有格点,所以答案就是\frac {q*(q-1)} 2​2​​q∗(q−1)​​.

个人想法：

其实自己就是画一个图，一定是方格图，然后自己画一下，就会发现一些规律，其实每个点一行一行的可以构成一个等差数列，所以相加得到的公式就是 (q-1)*(q-2)/2，注意 两个数相乘会爆long long，所以采用快速乘法 二进制那个给优化一下就搞定了:

My Code:

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
LL Mul_mod(LL a, LL b, LL m)
{
    LL ans = 0;
    while(b)
    {
        if(b & 1)
            ans = (ans+a)%m;
        b>>=1;
        a = (a+a)%m;
    }
    return ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        LL q, mod;
        cin>>q>>mod;
        LL x1 = q - 1;
        LL x2 = q - 2 ;
        if(x1%2==0)
            x1>>=1;
        else
            x2>>=1;
        printf("%lld\n",Mul_mod(x1,x2,mod));
    }
    return 0;
}