hdu 5666 Segment 俄罗斯乘法或者套大数板子

Segment

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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

题解：答案就是(（p-1)*(p-2)/2)%mod;p是10^18以内直接乘就会爆；利用俄罗斯乘法的加法性质得到答案；（现场做只会想到java大数。。果然菜鸟。。。）

　　　俄罗斯乘法：http://baike.baidu.com/link?url=vVo1zdml29g80N-BYvpdm2hNGpYwSnGoJsnAJmook4AJBiYUVL_ort5f7XqFJ0yx6zxB5ha90q6-1LD6HxPIaa

俄罗斯乘法代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define inf 2000000001
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF )  return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
ll eluosimul(ll x,ll y,ll mod)
{
    ll sum=0;
    while(x)
    {
        if(x&1)
        {
            sum+=y;
            sum%=mod;
        }
        x>>=1;
        y*=2;
        y%=mod;
    }
    return sum;
}
int main()
{
    ll x,y,z,i,t,m,q;
    scanf("%I64d",&x);
    while(x--)
    {
        ll ans;
        scanf("%I64d%I64d",&q,&m);
        if(q%2)
        ans=eluosimul(q-2,(q-1)/2,m);
        else
        ans=eluosimul(q-1,(q-2)/2,m);
        printf("%I64d
",ans);
    }
    return 0;
}

java：

import java.util.*;
import java.math.*;
public class Main {
    public static void main(String[] args) {
        Scanner cin=new Scanner(System.in);
        int x;
        x=cin.nextInt();
        while(x!=0)
        {
            x--;
            BigInteger c=new BigInteger("2");
            BigInteger e=new BigInteger("1");
            BigInteger a=cin.nextBigInteger();
            BigInteger d=a.subtract(c);
            BigInteger f=a.subtract(e);
            BigInteger b=cin.nextBigInteger();
            BigInteger ans=f.multiply(d);
            ans=ans.divide(c);
            System.out.println(ans.remainder(b));
        }
    }
}