poj2891

maksyuki 发表于 oj 分类，标签: 数论-线性同余方程(组)

15 9月 2015

Strange Way to Express Integers

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

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

题目类型：线性同余方程组

算法分析：直接使用线性同余方程组求解的方法计算即可

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>

#define lson rt << 1, l, m
#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int MOD = 7;
const int maxn = 100000 + 66;

long long ans[maxn], r[maxn];

long long n;


long long gcd_ex (long long a, long long b, long long &x, long long &y)
{
    if (b == 0) { x = 1; y = 0; return a; }
    long long d = gcd_ex (b, a % b, y, x);
    y = y - a / b * x;
    return d;
}



long long mod_equ_set ()
{
    long long M = ans[0], R = r[0], x, y, d;

    for (int i = 1; i < n; i++)
    {
        d = gcd_ex (M, ans[i], x, y);

        if ((r[i] - R) % d)
        return -1;

        x = (r[i] - R) / d * x % (ans[i] / d);
        R += x * M;
        M = M / d * ans[i];
        R %= M;
    }

return R > 0 ? R : R + M;
}

int main()
{
//    freopen ("aaa.txt", "r", stdin);

    while (scanf ("%lld", &n) != EOF)
    {
        int i;
        for (i = 0; i < n; i++)
            scanf ("%lld%lld", &ans[i], &r[i]);

        printf ("%lld\n", mod_equ_set ());
    }

return 0;
}

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

#define lson rt << 1, l, m

#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int MOD = 7;

const int maxn = 100000 + 66;

long long ans[maxn], r[maxn];

long long n;

long long gcd_ex (long long a, long long b, long long &x, long long &y)

{

if (b == 0) { x = 1; y = 0; return a; }

long long d = gcd_ex (b, a % b, y, x);

y = y - a / b * x;

return d;

}

long long mod_equ_set ()

{

long long M = ans[0], R = r[0], x, y, d;

for (int i = 1; i < n; i++)

{

d = gcd_ex (M, ans[i], x, y);

if ((r[i] - R) % d)

return -1;

x = (r[i] - R) / d * x % (ans[i] / d);

R += x * M;

M = M / d * ans[i];

R %= M;

}

return R > 0 ? R : R + M;

}

int main()

{

// freopen ("aaa.txt", "r", stdin);

while (scanf ("%lld", &n) != EOF)

{

int i;

for (i = 0; i < n; i++)

scanf ("%lld%lld", &ans[i], &r[i]);

printf ("%lld\n", mod_equ_set ());

}

return 0;

}

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