zoukankan      html  css  js  c++  java
  • 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 2q1018,1P1018,1T10.
     
    Output
        Output 1 number to each testcase,answer mod P.
     
    Sample Input
    1 2 107
     
    Sample Output
    0
     
    Source
    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;
    }
    View Code

    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));
            }
        }
    }
    View Code
  • 相关阅读:
    where T: class的解释
    调用钉钉的WebAPI接口实现与ERP数据的同步
    Json序列化和反序列化的方式
    Log4Net日志处理
    MVC项目中异常处理
    FindBI商业智能报表工具
    权限列表实现
    委托,匿名,lambda
    [经典贪心算法]贪心算法概述
    [zt]手把手教你写对拍程序(PASCAL)
  • 原文地址:https://www.cnblogs.com/jhz033/p/5400453.html
Copyright © 2011-2022 走看看