hdu 5666

题意：给出一个线段x+y= b，然后和0坐标组成的三角形内有多少整数点；

思路： 1 + …… + (n - 2)；等差数列求和（n-2）*(n - 1)/2;

运用大数相乘取模；

Segment

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1911    Accepted Submission(s): 713

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
  
  

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 100000 + 10;
//char ans[maxn];
ll mul(ll a,ll b,ll p)
{
	a %= p;
	b %= p; 
	ll ans = 0;
	ll t = a;
	while(b)
	{
		if(b & 1)
		{
			ans = (ans + t) % p;
		}
		b >>= 1;
		t = (t <<= 1) % p;
	}
	return ans;
}
ll n,p;
int main()
{
	int Tcase;
	scanf("%d",&Tcase);
	for(int ii = 1; ii <= Tcase; ii ++)
//	while( ~ scanf("%I64d%I64d",&n,&p))
	{
		scanf("%I64d%I64d",&n,&p);
		if(n < 0)
		n = -n;
		if(n - 2 < 1)
		{
			cout << 0 << endl;
			continue;
		}
		n -= 2;
		ll x,y;
		if((n + 1) % 2)
		{
			x = n / 2;
			y = (n + 1);
		}
		else 
		{
			x = (n + 1)/ 2;
			y = n;
		}
		cout << mul(x,y,p) << endl;
	}
	return 0;
}