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  • [POJ 2891] Strange Way to Express Integers

    Strange Way to Express Integers
    Time Limit: 1000MS   Memory Limit: 131072K
    Total Submissions: 10907   Accepted: 3336

    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 ≤ i ≤ k) 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 ≤ i ≤ k).

    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
     
    中国剩余定理、
    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    using namespace std;
    #define ll long long
    #define N 10010
    
    ll exgcd(ll a,ll b,ll& x, ll& y)
    {
        if(b==0)
        {
            x=1;
            y=0;
            return a;
        }
        ll d=exgcd(b,a%b,y,x);
        y-=a/b*x;
        return d;
    }
    bool solve(ll &m0,ll &a0,ll m,ll a)
    {
        ll y,x;
        ll g=exgcd(m0,m,x,y);
        if((a-a0)%g) return 0;
        x*=(a-a0)/g;
        x%=m/g;
        a0=(x*m0+a0);
        m0*=m/g;
        a0%=m0;
        if(a0<0) a0+=m0;
        return 1;
    }
    bool MLES(ll m[],ll r[],ll &m0 ,ll &a0,ll n)
    {
        m0=1;
        a0=0;
        bool flag=1;
        for(ll i=0;i<n;i++)
        {
            if(!solve(m0,a0,m[i],r[i]))
            {
                flag=0;
                break;
            }
        }
        return flag;
    }
    int main()
    {
        ll n;
        ll m[N],r[N];
        while(scanf("%lld",&n)!=EOF)
        {
            for(ll i=0;i<n;i++)
            {
                scanf("%lld%lld",&m[i],&r[i]);
            }
            ll m0,a0;
            ll flag=MLES(m,r,m0,a0,n);
            if(flag) printf("%lld
    ",a0);
            else printf("-1
    ");
        }
        return 0;
    }
    趁着还有梦想、将AC进行到底~~~by 452181625
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  • 原文地址:https://www.cnblogs.com/hate13/p/4442977.html
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