s11

\documentclass[a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage{url}
\usepackage{minted}
\usepackage{graphicx}

\title{UAP Training}
\author{Rafid Bin Mostofa}
\date{July 13, 2019}

\begin{document}

\maketitle
\hline
\tableofcontents
\pagebreak

\section{Chinese Remainder Theorem}
This is a good source: \url{https://cp-algorithms.com/algebra/chinese-remainder-theorem.html} \\
\subsection{Problems}
\begin{itemize}
    \item \url{http://www.lightoj.com/volume_showproblem.php?problem=1319} [\ref{code:monkey}]
    \item RUET 2019 A
\end{itemize}

\section{Flow}

\subsection{Maximum Flow}
\begin{figure}[h]
    \centering
    \includegraphics{maxflow}
    \label{fig:my_label}
\end{figure}


\begin{itemize}
    \item \url{https://www.topcoder.com/community/competitive-programming/tutorials/maximum-flow-section-1/}
    \item \url{https://www.topcoder.com/community/competitive-programming/tutorials/maximum-flow-section-2/}
\end{itemize}
Problems: \url{http://www.lightoj.com/volume_problemcategory.php?user_id=29559&category=Max%20Flow/Min%20Cut}

\subsection{Min Cost Max Flow}
Source: \url{https://cp-algorithms.com/graph/min_cost_flow.html} \\ \\
Problems: \url{http://www.lightoj.com/volume_problemcategory.php?user_id=29559&category=Min%20Cost%20Max%20Flow} \\
Another Problem: \url{https://toph.co/p/alice-loves-to-play-game}

\subsection{Bipartite Matching}
\begin{itemize}
    \item Hopcroft Karp
    \item Kuhn
    \item Hungarian (Weighted)
\end{itemize}

\pagebreak
\section{APPENDIX}
\subsection{LightOJ 1319 - Monkey Tradition}
\label{code:monkey}

\begin{minted}{cpp}
#include <bits/stdc++.h>
using namespace std;

typedef pair<long long, long long> pll;
#define ff first
#define ss second

pll extendedEuclid(int a, int b) { // returns x, y | ax + by = gcd(a,b)
    if(b == 0) return pll(1, 0);
    else {
        pll d = extendedEuclid(b, a % b);
        return pll(d.ss, d.ff - d.ss * (a / b));
    }
}

long long modInv(int a, long long p) {
    pll ret = extendedEuclid(a, p);
    return ((ret.ff % p) + p) % p;
}

unsigned long long crt(vector<int> &rem, vector<int> &mod) {
    assert(rem.size() == mod.size());

    unsigned long long ret = 0;
    int n = rem.size();

    unsigned long long MOD = 1;
    for(int i=0; i<n; ++i) MOD *= mod[i];

    // you shouldn't run a n^2 loop inside crt
    // this case was exceptional and
    // the programmer was too lazy to optimize
    for(int i=0; i<n; ++i) {
        long long cur = rem[i];
        for(int j=0; j<n; ++j) {
            if(i == j) continue;
            cur = (cur * mod[j]);
            cur = (cur * modInv(mod[j], mod[i]));
            cur %= MOD;
        }
        ret = (ret + cur) % MOD;
    }

    return ret;
}

int main() {
    int t, tc=0;
    scanf("%d", &t);

    while(t--) {
        int n;
        scanf("%d", &n);
        vector<int> rem(n), mod(n);
        for(int i=0; i<n; ++i) cin >> mod[i] >> rem[i];

        unsigned long long res = crt(rem, mod);
        printf("Case %d: %llu\n", ++tc, res);
    }

    return 0;
}
\end{minted}
\end{document}