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  • 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
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  • 原文地址:https://www.cnblogs.com/jhz033/p/5400453.html
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