FZU 1402 猪的安家 中国剩余定理

http://acm.fzu.edu.cn/problem.php?pid=1402

逗比题..和前面那题一样解就行了...

反正都是素数,就把中国剩余定理拓展一下...普及姿势好了:

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逆元:

对于同余方程 :

 

我们有 : 

x就是A对B的逆元了 用EX_GCD求出x即可 ,这里x可能为负数, 对B做正取模

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中国剩余定理 :

对于方程组

我们有如下性质

然后带公式即可

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/********************* 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        1005
#define MOD         (int)1e9+7
#define PI          acos(-1.0)
#define DINF        (1e10)
#define LINF        ((1LL)<<50)
#define INF         (0x3f3f3f3f)
#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   ************************/

pair<LL,LL> ex_gcd(LL a,LL b){
    if(b == 0) return make_pair(1,0);
    pair<LL,LL> t = ex_gcd(b,a%b);
    return make_pair(t.second , t.first - (a / b) * t.second);
}
LL China_Remainder(LL a[],LL b[],int n){
    LL res = 0,det = 1;
    for(int i = 0 ; i < n ; i++) det *= a[i];
    for(int i = 0 ; i < n ; i++){
        LL xx = det / a[i];
        pair<LL,LL> t = ex_gcd(xx,a[i]);
        t.first = (t.first % a[i] + a[i]) % a[i]; // 求xx对a[i]的逆元 ,最后对a[i]取模
        res = (res + (xx * b[i] * t.first) % det) % det;
    }
    return res;
}
int main()
{
    int n;
    while(cin>>n){
        LL a[20],b[20];
        for(int i = 0 ; i < n ; i++){
            cin>>a[i]>>b[i];
        }
        LL ans = China_Remainder(a,b,n);
        cout<<ans<<endl;
    }
    return 0;
}