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chinese_remainder_theorem.cpp
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chinese_remainder_theorem.cpp
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#include <cstdio>
#include <iostream>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <math.h>
#include <cmath>
#include <string>
#include <iomanip>
#include <cstring>
#include <sstream>
#include <algorithm>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define inf 1013161010
#define ninf -1013161010
#define mod 1000000007
#define ll long long
#define lf long double
#define in(x) scanf("%d",&x);
#define sz(x) ((int)x.size())
#define lld l64d
#define rep(i,n) for(i=0;i<n;i++)
#define rrep(i,n) for(i=n-1;i>=0;i--)
#define rep1(i,a,b) for(i=a;i<=b;i++)
#define rrep1(i,a,b) for(i=a;i>=b;i--)
#define stlfor(i,t) for(auto i =t.begin();i!=t.end();i++)
#define fr freopen("x.txt","r",stdin)
#define frc freopen("y.txt","w",stdout)
#define all(x) x.begin(),x.end()
#define set0(x) memset(x,0,sizeof(x))
#define dbg cout<<"yo "<<endl;
#define pset(n) fixed<<showpoint<<setprecision(n)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vpii vector<pair<int,int> >
#define vpll vector<pair<ll,ll> >
#define si set<int>
#define mii map<int,int>
#define umii unordered_map<int,int>
#define vi vector<int>
#define pb push_back
#define ff first
#define ss second
#define mp make_pair
typedef tree<pii,null_type,less<pii>,rb_tree_tag,tree_order_statistics_node_update> OST;
ll toint(const string &s) { stringstream ss; ss << s; ll x; ss >> x; return x; }
string tostring ( ll number ){ stringstream ss; ss<< number; return ss.str();}
const lf pi = 2*acos(0);
const int nn = 2000006;
ll gcd(ll a,ll b){return (b==0)? a:gcd(b,a%b); }
void nope(int num = 0){ if(num==0) cout<<"NO"; else cout<<"-1"; exit(0); }
ll x, y, d;
void egcd(ll a, ll b)
{
if(b==0)
{
x = 1;
y = 0;
d = a;
return ;
}
egcd(b, a%b);
ll x1 = y;
ll y1 = x - (a/b)*y;
x = x1;
y = y1;
}
// THIS SOLVES CRT FOR CO-PRIME MODULI
ll a[100], b[100];
int main()
{
ios_base::sync_with_stdio(false); cin.tie(0);
ll i,j,n,tt;
tt=1;
//cin>>tt;
while(tt--)
{
cin>>n;
// Problem :
// given n equations as follows:
// x = a[i] (mod b[i])
// find x
ll N = 1, z;
rep(i,n)
{
cin>>a[i]>>b[i];
N*=b[i];
}
ll ans = 0;
for(i=0;i<n;i++)
{
z = N/b[i];
egcd(z, b[i]);
x = (x+b[i])%b[i];
ans+=(a[i]*z*x);
ans = (ans+N)%N;
}
cout<<ans<<endl;
}
return 0;
}