poj

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 14547		Accepted: 4759

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.

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.

Source

POJ Monthly--2006.07.30, Static

题意：
给你n个线性同余方程，让你求解这个方程组。
思路：
对于同余方程(x≡b(mod m) 等价于 x = b1 + m1*x1；所以对于两个同余方程组，就有b1+m1*x1=b2+m2*x2。移项得到m2*y2-m1*y1=b1-b2；根据扩展欧几里得课以求出y1的值，带回就可以求出X,X是这个方程的一个特解，通解为X = X'+k*lcm(m1, m2),两边同时取模，得到X mod lcm（m1. m2）= X'，这就是两个方程合并之后的结果。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const long long MOD = 1e9 + 7;
const int MAXN = 1e5;
void ext_gcd(LL a, LL b, LL& d, LL& x, LL& y) {
    if (!b){x = 1;d = a; y = 0;}
    else {
        ext_gcd(b, a%b, d, y, x);
        y -= x*(a/b);
    }
}
void solved(int n) {
    bool flag = false;
    LL a0, r0, a1, r1, x, y, a ,b, c, d;
    scanf("%lld%lld", &a0, &r0);
    for (int i = 1; i < n; i++) {
        scanf("%lld%lld", &a1, &r1);
        a = a0, b = a1, c = r1 - r0;
        ext_gcd(a, b, d, x, y);
        if (c%d) flag = true;
        x = ((x*c/d)%(b/d) + b/d)%(b/d);
        r0 = a0*x + r0;
        a0 = a1*a0/d;
    }
    if (flag) printf("-1
");
    else printf("%lld
", r0);
}
int main() {
    int n;
    while (scanf("%d", &n) != EOF) {
        solved(n);
    }
    return 0;
}