POJ 2981 Strange Way to Express Integers 模线性方程组

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

结果看了半天还是没懂那个模的含义...懂了我再补充...

其他的思路都在注释里

/********************* Template ************************/
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;

#define EPS         1e-8
#define MAXN        (int)1e5+100
#define MOD         (int)1e9+7
#define PI          acos(-1.0)
#define LINF        ((1LL)<<50)
#define INF            (1<<30);
#define max(a,b)    ((a) > (b) ? (a) : (b))
#define min(a,b)    ((a) < (b) ? (a) : (b))
#define max3(a,b,c) (max(max(a,b),c))
#define min3(a,b,c) (min(min(a,b),c))
#define BUG         cout<<"BUG! "<<endl
#define LINE        cout<<"--------------"<<endl
#define L(t)        (t << 1)
#define R(t)        (t << 1 | 1)
#define Mid(a,b)    ((a + b) >> 1)
#define lowbit(a)   (a & -a)
#define FIN            freopen("in.txt","r",stdin)
#define FOUT        freopen("out.txt","w",stdout)
#pragma comment     (linker,"/STACK:102400000,102400000")

typedef long long LL;
// typedef unsigned long long ULL;
// typedef __int64 LL;
// typedef unisigned __int64 ULL;
LL gcd(LL a,LL b){ return b?gcd(b,a%b):a; }
LL lcm(LL a,LL b){ return a/gcd(a,b)*b; }

/*********************   F   ************************/

LL a[MAXN],r[MAXN];
bool flag;

pair<LL,LL> ex_gcd(LL a,LL b){
    if(b == ) return make_pair(,);
    pair<LL,LL> t = ex_gcd(b,a%b);
    return make_pair(t.second , t.first - (a / b) * t.second);
}

LL work(int n){
    LL a0 = a[],r0 = r[];
    LL tmp,agcd,pr;
    for(int i =  ; i < n ; i++){
        pair<LL,LL> p = ex_gcd(a0,a[i]);
        agcd = gcd(a0,a[i]);
        pr = r[i] - r0;
        if(pr % agcd) {             // pr%agcd==0 保证有解
            flag = true;
            return ;
        }
        // 不明这个模的意义,本来是要%a[i]的现在 放大了(pr/agcd)倍,估计是/pr求逆元的思想吧
        tmp = a[i] / agcd;
        //还原方程 : p.first*a0≡pr(mod a[i])
        p.first = (pr / agcd * p.first % tmp + tmp) % tmp;
        r0 = r0 + a0 * p.first;     // 满足两个方程最小整数
        a0 = a0 / agcd * a[i] ;     // a0=LCM(a0,a[i]) 保证解的最小...具体为什么本弱说不清
    }
    return r0;
}
int main()
{
    //FIN;
    //FOUT
    int n;
    while(cin>>n){
        flag = false;
        for(int i =  ; i < n ; i++)
            cin>>a[i]>>r[i];
        LL ans = work(n);
        if(flag) cout<<"-1"<<endl;
        else cout<<ans<<endl;
    }
    return ;
}