【POJ 2891】 Strange Way to Express Integers

【题目链接】

http://poj.org/problem?id=2891

【算法】

exgcd

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
typedef long long ll;

int i,k;
ll a[],m[];

inline ll exgcd(ll a,ll b,ll &x,ll &y)
{
        ll g;
        if (b == )
        {
                x = ;
                y = ;
                return a;
        } else
        {
                g = exgcd(b,a%b,y,x);
                y -= a / b * x;
                return g;
        }
}
inline ll solve()
{
        int i;
        ll g,x,y,M = a[],R = m[];
        for (i = ; i <= k; i++)
        {
                g = exgcd(M,a[i],x,y);
                if ((R - m[i]) % g != ) return -;
                x = (R - m[i]) / g * x % a[i];
                R -= x * M;
                M = M / g * a[i];
                R %= M;
        }
        return (R % M + M) % M;
}
int main()
{

        while (scanf("%d",&k) != EOF)
        {
                for (i = ; i <= k; i++) scanf("%lld%lld",&a[i],&m[i]);
                printf("%lld\n",solve());
        }
        return ;

}