zoukankan      html  css  js  c++  java
  • Strange Way to Express Integers(中国剩余定理+不互质)

    Strange Way to Express Integers

    Time Limit:1000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I64u
     

    Description

    Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:

    Choose k different positive integers a1, a2, …, ak. For some non-negative m, divide it by every ai (1 ≤ ik) to find the remainder ri. If a1, a2, …, ak are properly chosen, m can be determined, then the pairs (ai, ri) can be used to express m.

    “It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”

    Since Elina is new to programming, this problem is too difficult for her. Can you help her?

    Input

    The input contains multiple test cases. Each test cases consists of some lines.

    • Line 1: Contains the integer k.
    • Lines 2 ~ k + 1: Each contains a pair of integers ai, ri (1 ≤ ik).

    Output

    Output the non-negative integer m on a separate line for each test case. If there are multiple possible values, output the smallest one. If there are no possible values, output -1.

    Sample Input

    2
    8 7
    11 9

    Sample Output

    31

    Hint

    All integers in the input and the output are non-negative and can be represented by 64-bit integral types.

    题意:给你k组数。x%M[i]=A[i];

    思路:中国剩余定理,扩展欧几里德

    不会的可以参考:http://blog.csdn.net/u010579068/article/details/45422941

    转载请注明出处:寻找&星空の孩子

    题目链接:http://poj.org/problem?id=2891

    #include<stdio.h>
    #define LL __int64
    
    void exgcd(LL a,LL b,LL& d,LL& x,LL& y)
    {
        if(!b){d=a;x=1;y=0;}
        else
        {
            exgcd(b,a%b,d,y,x);
            y-=x*(a/b);
        }
    }
    LL gcd(LL a,LL b)
    {
        if(!b){return a;}
        gcd(b,a%b);
    }
    
    LL M[55000],A[55000];
    
    
    LL China(int r)
    {
        LL dm,i,a,b,x,y,d;
        LL c,c1,c2;
        a=M[0];
        c1=A[0];
        for(i=1; i<r; i++)
        {
            b=M[i];
            c2=A[i];
            exgcd(a,b,d,x,y);
            c=c2-c1;
            if(c%d) return -1;//c一定是d的倍数,如果不是,则,肯定无解
            dm=b/d;
            x=((x*(c/d))%dm+dm)%dm;//保证x为最小正数//c/dm是余数,系数扩大余数被
            c1=a*x+c1;
            a=a*dm;
        }
        if(c1==0)//余数为0,说明M[]是等比数列。且余数都为0
        {
            c1=1;
            for(i=0;i<r;i++)
                c1=c1*M[i]/gcd(c1,M[i]);
        }
        return c1;
    }
    int main()
    {
        int n;
    
        while(scanf("%d",&n)!=EOF)
        {
            for(int i=0;i<n;i++)
            {
                scanf("%I64d%I64d",&M[i],&A[i]);
            }
            if(n==1){ printf("%I64d
    ",A[0]);continue;}
            LL ans=China(n);
            printf("%I64d
    ",ans);
    
        }
        return 0;
    }
  • 相关阅读:
    ArcGIS for Android地图控件的5大常见操作
    adb开启不了解决方案
    Eclipse中通过Android模拟器调用OpenGL ES2.0函数操作步骤
    解决 Your project contains error(s),please fix them before running your application问题
    二路归并算法实现
    字符串全排列
    python连接MySQL
    .net常考面试题
    win7 web开发遇到的问题-由于权限不足而无法读取配置文件,无法访问请求的页面
    int.Parse()与int.TryParse()
  • 原文地址:https://www.cnblogs.com/yuyixingkong/p/4472790.html
Copyright © 2011-2022 走看看