HDOJ-----5666---Segment---加法快速幂

最新推荐文章于 2020-08-26 12:51:29 发布

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最新推荐文章于 2020-08-26 12:51:29 发布

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分类专栏：快速幂与同余

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本文介绍了一个计算特定直角三角形内不在边界的整数点数量的方法。通过数学推导，给出了一个有效算法，该算法利用快速幂运算求解答案模P的问题。

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Segment

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1815 Accepted Submission(s): 669

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

x+y=p这条直线与坐标轴围成的三角形中不在边界上的整数点有多少个

从x轴开始，向上或者向下（p的正负），依次是p-2， p-3， p-4 …… 1，一共是（p-1）*（p-2）/2个

#include<cstdio>
#include<iostream>
#define LL long long
using namespace std;
LL solve(LL a, LL b, LL c){
    LL ans = 0;
    while(b){//加法快速幂，和乘法基本一样
        if(b&1) ans = (ans+a)%c;
        a = 2*a%c;
        b >>= 1;
    }
    return ans;
}
int main(){
    int t;
    LL q, p;
    cin >> t;
    while(t--){
        cin >> q >> p;
        if((q-1)&1) cout << solve((q-2)/2, q-1, p) << endl;//考虑到除2是否丢精度
        else cout << solve((q-1)/2, q-2, p) << endl;
    }
    return 0;
}