记录编号	570333	评测结果	AAAAAAAAAA
题目名称	[POJ 2891]表达整数的奇怪方式	最终得分	100
用户昵称	lihaoze	是否通过	通过
代码语言	C++	运行时间	0.000 s
提交时间	2022-03-27 23:55:08	内存使用	0.00 MiB

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <string>
#define OPEN(_x) freopen(#_x".in", "r", stdin); freopen(#_x".out", "w", stdout)
#define MAX(_a, _b) [&](int __a, int __b) { return __a < __b ? __b : __a; }((_a), (_b))
#define MIN(_a, _b) [&](int __a, int __b) { return __a > __b ? __b : __a; }((_a), (_b))
#define ABS(_x) [&](int __x) { return __x < 0 ? -__x : __x; }(_x)
#define fi first
#define se second 

using ll = long long;
using PII = std::pair<int, int>;

namespace IO {
    template <typename T> inline T read() {
        char ch = getchar();
        T ret = 0, sig = 1;
        while(ch < '0' || ch > '9') { if(ch == '-') sig = -1; ch = getchar(); }
        while(ch >= '0' && ch <= '9') ret *= 10, ret += ch - 48, ch = getchar();
        return ret * sig;
    }
    template <typename T> inline void write(T out) {
        if(!out) { putchar('0'), putchar(' '); return; }
        int stk[100], tt = 0;
        if(out < 0) out = -out, putchar('-');
        while(out) stk[tt++] = out % 10, out /= 10;
        for(register int i = --tt; i>=0; --i) putchar(stk[i] + 48);
        putchar(' ');
    }
    template <typename T> inline void read(T& ret) { ret = IO::read<T>(); }
    template <typename T, typename... Args> inline void read(T& x, Args&... args) { IO::read(x), IO::read(args...); }
    template <typename T, typename... Args> inline void write(T x, Args... args)  { IO::write(x), IO::write(args...); }
};


const int N = 30;
int n;

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

int main() {
    #ifdef DEBUG
    OPEN(test);
    #endif
    OPEN(strangeway);
    IO::read(n);
    ll a1, m1, a2, m2;
    IO::read(a1, m1);
    for (register int i = 1; i<=n - 1; ++i) {
        IO::read(a2, m2);
        ll k1, k2;
        ll d = exgcd(a1, a2, k1, k2);
        if ((m2 - m1) % d) {
            puts("-1");
            return 0;
        }
        k1 *= (m2 - m1) / d;
        ll t = a2 / d;
        k1 = (k1 % t + t) % t;

        m1 += a1 * k1;
        a1 = std::abs(a1 / d * a2);
    }
    IO::write((m1 % a1 + a1) % a1);
    return 0;
}