POJ 2891 Strange Way to Express Integers（拓展欧几里得）

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, …, a_k. For some non-negative m, divide it by every a_i (1 ≤ i ≤ k) to find the remainder r_i. If a₁, a₂, …, a_k are properly chosen, m can be determined, then the pairs (a_i, r_i) 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 a_i, r_i (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.

 

题目大意：给k个线性同余方程，求这些方程的公共解。

附思路：http://blog.csdn.net/orpinex/article/details/6972654

思路：考虑两个方程的情况

ans = r₁(mod a₁)
ans = r₂(mod a₂)①
存在k₁使得ans = r₁ + k₁ * a₁
把ans代入①得：r₁ + k₁ * a₁ = r₂(mod a₂)
k₁ * a₁ = r₂ - r₁(mod a₂)
存在k₂使得k₁ * a₁ - k₂ * a₂ = r₂ - r₁
利用拓展欧几里得求出k₁（为了得到最小的非负整数k₁，可以让k₁ = k₁ mod (a₂/gcd(a₁, a₂))）

那么令ans = r₁ + k₁ * a₁

对于多个方程的情况，两个两个地联立解，new_r = ans = r₁ + k₁ * a₁, new_a = lcm(a₁, a₂)

解析：

关于为了让k₁最小要k₁ = k₁ mod (a₂/gcd(a₁, a₂))。令d = gcd(a₁, a₂)，a₁'= a₁ / d，a₂' = a₂ / d ，r' = (r₂ - r₁) / d，对方程k₁ * a₁ - k₂ * a₂ = r₂ - r₁，两边同时除以d得k₁ * a₁' - k₂ * a₂' = r'，即k₁ * a₁' = r' (mod a₂')，对于任意解k₁ = x'，有通解x = x' + a₂' = x' + a₂ / gcd(a₁, a₂)。则最小的k1 = (x' + a₂ / gcd(a₁, a₂)) mod (a₂ / gcd(a₁, a₂) = x' mod (a₂ / gcd(a₁, a₂))

关于new_a = lcm(a₁, a₂)。考虑方程x * a = y * b，两边同时除以gcd(a, b)，得到x * a' = y * b'，x / y = b' / a'，那么有x = kb'，y = ka'，k∈Z。那么使得方程x * a = y * b成立的x * a = k * b' * a = k * lcm(a, b)。容易想象，p + ax = q + by的每个合理的p + ax的差为lcm(a, b)。即对于方程r₁(mod a₁) = r₂(mod a₂)的解也是隔lcm(a₁, a₂)就出现一个解，即new_a = lcm(a₁, a₂)。 

 

代码（16MS）：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;

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);
    }
}

int main() {
    LL k, a1, a2, r1, r2;
    while(scanf("%I64d", &k) != EOF) {
        bool flag = true;
        scanf("%I64d%I64d", &a1, &r1);
        for(int i = 1; i < k; ++i) {
            scanf("%I64d%I64d", &a2, &r2);
            if(!flag) continue;
            LL r = r2 - r1, d, k1, k2;
            exgcd(a1, a2, d, k1, k2);
            if(r % d) flag = false;
            LL t = a2 / d;
            k1 = (r / d * k1 % t + t) % t;
            r1 = r1 + a1 * k1;
            a1 = a1 / d * a2;
        }
        printf("%I64d\n", flag ? r1 : -1);
    }
}