Problem:
Score : 100 points

Problem Statement

We have N+M balls, each of which has an integer written on it.

It is known that:

The numbers written on N of the balls are even.

The numbers written on M of the balls are odd.

Find the number of ways to choose two of the N+M balls (disregarding order) so that the sum of the numbers written on them is even.

It can be shown that this count does not depend on the actual values written on the balls.

Constraints

0 \leq N,M \leq 100

2 \leq N+M

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the answer.

Sample Input 1

2 1

Sample Output 1

1

For example, let us assume that the numbers written on the three balls are 1,2,4.

If we choose the two balls with 1 and 2, the sum is odd;

If we choose the two balls with 1 and 4, the sum is odd;

If we choose the two balls with 2 and 4, the sum is even.

Thus, the answer is 1.

Sample Input 2

4 3

Sample Output 2

9

Sample Input 3

1 1

Sample Output 3

0

Sample Input 4

13 3

Sample Output 4

81

Sample Input 5

0 3

Sample Output 5

3

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using static System.Console;
using static System.Math;
 
public class Hello{
    static public int mod = 1000000007;
    public static void Main(){
        // 方針
        // 
        int[] aa = inta();
        int a = aa[0];
        int b = aa[1];
        a = a*(a-1)/2;
        b = b*(b-1)/2;
        WriteLine(a+b);
        
        
        
        
        
        
        
        
        
        
        
    }
    static int ncr(int n,int r){r = Min(r,n-r);long nn = 1;for(int i=n-r+1;i<=n;i++){nn = nn*i%mod;}long rr = 1;for(int i=1;i<=r;i++){rr = rr*i%mod;}rr = square((int)rr,mod-2);nn = nn * rr %mod;return (int)nn;}
    // a^b mod
    static int square(int a,int b){string binary = Convert.ToString(b,2);int bileng = binary.Length;long a_power = a;long value = 1;for(int i=bileng-1;i>=0;i--){if(binary[i] == '1'){value = value*a_power%mod;}a_power = a_power*a_power%mod;}return (int)value;}
    //static int square2(long a,long b,long m){var list = new List<int>();int temp = 1;long n2 = a;list.Add(1);int count = 1;while(temp < b){list.Add((int)n2);temp*=2;count++;n2 = n2*n2%m;}long ret = 1;while(b > 0){if(temp <= b){b -= temp;ret = ret*list[count]%1000000007;}count--;temp/=2;}return(int)ret;}
    //各種読取
    static string rstr(){ return ReadLine(); }
    static int rint(){ return int.Parse(ReadLine()); }
    static long rlong(){ return long.Parse(ReadLine()); }
    static string[] stra(){ return ReadLine().Split(' '); }
    static int[] inta(){ string[] read_str_array = ReadLine().Split(' '); int[] return_int_array = new int[read_str_array.Length]; for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_int_array[countup_i] = int.Parse(read_str_array[countup_i]); } return return_int_array; }
    static int[,] inta2(int num_array,int in_array){ int[,] int_array2 = new int[num_array,in_array]; for(int i=0;i<num_array;i++){ int[] temp_array = inta(); for(int j=0;j<in_array;j++){ int_array2[i,j] = temp_array[j]; } } return int_array2; }
    static long[] longa(){ string[] read_str_array = ReadLine().Split(' '); long[] return_long_array = new long[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_long_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_long_array; }
    static double[] doublea(){ string[] read_str_array = ReadLine().Split(' '); double[] return_double_array = new double[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_double_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_double_array; }
    // -----------------------------
    static long divideup(long divided,long divid){ long temp_divide = divided/divid; if(divided % divid > 0){ temp_divide++; } return temp_divide; }
    static long GCD(long a,long b){ if(a < b){ long temp = a; a = b; b = temp; } if(a % b == 0){ return b; } else{ long temp = b; b = a%b; a = temp; return GCD(a,b); } }
    static long LCM(long a,long b){ return a * b / GCD(a,b); }
}



//知見
// (string) hogestr -> (char[]) hoge
    // char hoge = hogestr.ToCharArray();
// (char[]) hoge -> (string) hogestr
    // string hoge = new string(hoge);
// Hashtable ×  Dictionary ○
// Dictionary hoge<string,int>
    // hoge.ContainsKey(string) -> return boolean
    // hoge[string] -> return int
    // hoge[int] -> CE
// Dictionary hoge<int1,int2>
    // hoge<int1,int2>: hoge[int1] -> return int2
    // KeyValuePair<int,int> k <-> k.Key == int1 / k.Value == int2
// SortedDictionary() > SortedList() ???
// キュー　-> 幅優先 ???
    // var queue = new Queue<int>(); queue.Enqueue() queue.Dequeue()
// スタック -> 深さ優先 ???
    // var stack = new Stack<int>(); stack.Push(hoge) stack.Pop() stackPeek() stack.Count
// if (list?.Count > 0) リストが空でもnullでもない
// ソート
    // list.Sort(StringComparer.Ordinal); listの(辞書順?)ソート
        // StringComparer.Ordinal とは
// 出力
    // foreach(var a int list){ WriteLine(a); } より
        // WriteLine(string.Join("\n", list)); が速い
    //var sw = new StreamWriter(Console.OpenStandardOutput()){AutoFlush = false};
    //    Console.SetOut(sw);
    //    foreach(var k in list[num-1]){
    //        WriteLine(k);
    //    }
    //    Console.Out.Flush();