poj2891 Strange Way to Express Integers

Strange Way to Express Integers

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 17425		Accepted: 5863

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

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

分析：模板.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
ll a[1010], b[1010], n;

ll gcd(ll a, ll b)
{
    if (!b)
        return a;
    return gcd(b, a % b);
}

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (!b)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll temp = exgcd(b, a % b, x, y), t = x;
    x = y;
    y = t - (a / b) * y;
    return temp;
}

ll niyuan(ll x, ll mod)
{
    ll px, py, t;
    t = exgcd(x, mod, px, py);
    if (t != 1)
        return -1;
    return (px % mod + mod) % mod;
}

bool hebing(ll a1, ll n1, ll a2, ll n2, ll &a3, ll &n3)
{
    ll d = gcd(n1, n2), c = a2 - a1;
    if (c % d != 0)
        return false;
    c = (c % n2 + n2) % n2;
    n1 /= d;
    n2 /= d;
    c /= d;
    c *= niyuan(n1, n2);
    c %= n2; //取模,在哪一个模数下就要模哪个,模数要跟着变化.
    c *= n1 * d;
    c += a1;
    n3 = n1 * n2 * d;
    a3 = (c % n3 + n3) % n3;
    return true;
}

ll China()
{
    ll a1 = b[1], n1 = a[1], a2, n2;
    for (int i = 2; i <= n; i++)
    {
        ll a3, n3;
        a2 = b[i], n2 = a[i];
        if (!hebing(a1, n1, a2, n2, a3, n3))
            return -1;
        a1 = a3;
        n1 = n3;
    }
    return (a1 % n1 + n1) % n1;
}

int main()
{
    scanf("%lld", &n);
    for (int i = 1; i <= n; i++)
        scanf("%lld%lld", &a[i], &b[i]);
    printf("%lld
", China());

    return 0;
}