hdu 5666 Segment(快速乘)

最新推荐文章于 2020-02-23 23:52:03 发布

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最新推荐文章于 2020-02-23 23:52:03 发布

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分类专栏： HDU 杂题快速幂/乘

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本文介绍了一个有趣的平面几何问题：给定一条直线x+y=q，该直线与坐标轴相交形成一个三角形区域。任务是计算在这个三角形区域内且不在边界上的整数坐标点的数量，并输出结果对P取模后的值。文章提供了两种编程解决方案，一种使用C++实现快速乘法，另一种使用Java实现。

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

solution：
推出公式为ans=(q-1)*(q-2)%p;但因为乘法会爆longlong，因此可用快速乘或Java
快速乘：
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll p, q;
ll pow(ll x,ll y)
{
    ll ans = 0;
    while (y)
    {
        if (y%2==1)ans=(ans+x)%p;
        x =(x+x)%p;
        y >>= 1;
    }
    return ans%p;
}
int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%I64d%I64d", &q, &p);
        if (q%2==1)printf("%I64d\n", pow(q - 2, (q - 1) / 2));
        else printf("%I64d\n", pow(q - 1, (q - 2) / 2));
    }
    return 0;
}

Java:
import java.math.BigInteger;
import java.util.*;
public class Main {
    public static void main(String[] args) {
        Scanner cin=new Scanner(System.in);
        BigInteger p,q,mod;
        BigInteger one=new BigInteger("1");
        BigInteger two=new BigInteger("2");
        int T;T=cin.nextInt();
        for(int i=0;i<T;i++)
        {
            q=cin.nextBigInteger();
            p=cin.nextBigInteger();
            System.out.println((((q.subtract(one)).multiply(q.subtract(two))).divide(two)).mod(p));
        }
    }
}