Problem:
Score: 100 points

Problem Statement

A triple of numbers is said to be poor when two of those numbers are equal but the other number is different from those two numbers.

You will be given three integers A, B, and C. If this triple is poor, print Yes; otherwise, print No.

Constraints

A, B, and C are all integers between 1 and 9 (inclusive).

Input

Input is given from Standard Input in the following format:

A B C

Output

If the given triple is poor, print Yes; otherwise, print No.

Sample Input 1

5 7 5

Sample Output 1

Yes

A and C are equal, but B is different from those two numbers, so this triple is poor.

Sample Input 2

4 4 4

Sample Output 2

No

A, B, and C are all equal, so this triple is not poor.

Sample Input 3

4 9 6

Sample Output 3

No

Sample Input 4

3 3 4

Sample Output 4

Yes

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#include<stdio.h>
#include<sys/types.h>
#include<sys/stat.h>
#include<fcntl.h>
#include<unistd.h>
#include<stdlib.h>

using namespace std;

typedef unsigned long long ull;
typedef long long int ll;
typedef long double ld;
#define REP(i,a,b) for(ll i=a;i<=b;++i)
#define REV(i,a,b) for(ll i=a;i>=b;i--)
#define pll pair<ll,ll>
#define vll vector<ll>
#define sll set<ll>
#define vpll vector<pll>
#define F first
#define S second
#define PB push_back
#define MP make_pair
#define min_pq priority_queue <ll, vector<ll>, greater<ll> >
#define max_pq priority_queue <ll>
#define deci(n) fixed << setprecision(n)
#define INF 1e18

ll mod=1000000007;
////__RANGE__UPDATE___///
vector<ll> initializeDiffArray(vector<ll>& A);
void update(vector<ll>& D, ll l, ll r, ll x);
void getArray(vector<ll>& A, vector<ll>& D);
////__RANGE__UPDATE___///
ll min(ll a,ll b);
ll max(ll a,ll b);
ll gcd(ll a, ll b);
void swap(ll *a,ll *b);
ll lcm(ll a, ll b);
ll modpower(ll x, ll y, ll p);

//Only for integer y's
ll power(ll x, ll y);
ll modulo( ll value, ll m);
ll myXOR(ll x, ll y) ;
ll diff(ll a,ll b);


int main()
{
  /*  #ifdef DEBUG
    freopen("std.in", "r", stdin);
    freopen("std.out", "w", stdout);
    #else
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif */
ll rep=1;
//cin>>rep;
while(rep--)
{
ll a[3];cin>>a[0]>>a[1]>>a[2];sort(a,a+3);
if(a[0]==a[1]&&a[0]!=a[2])
cout<<"Yes"<<'\n';
else
if(a[1]==a[2]&&a[1]!=a[0])
cout<<"Yes"<<'\n';
else
cout<<"No"<<'\n';
}

}






ll myXOR(ll x, ll y)
{
   return (x | y) & (~x | ~y);
}

ll min(ll a,ll b)
{
  if(a>b)
  return b;
  return a;
}

ll max(ll a,ll b)
{
  if(a<b)
  return b;
  return a;
}

ll gcd(ll a, ll b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

ll diff(ll a,ll b)
{
  if(a>b)
  return a-b;
  return b-a;
}

void swap(ll *a,ll *b)
{
  ll t=*a;
  *a=*b;
  *b=t;
}

ll lcm(ll a, ll b)
{
    if(a==0)
    return b;
    if(b==0)
    return a;
    return (a*b)/gcd(a, b);
}

ll modpower(ll x, ll y, ll p)
{
    ll res = 1;      // Initialize result

    x = x % p;  // Update x if it is more than or
                // equal to p

    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % p;

        // y must be even now
        y = y>>1; // y = y/2
        x = (x*x) % p;
    }
    return res;
}

ll power(ll x, ll y)
{
    ll temp;
    if( y == 0)
        return 1;
    temp = power(x, y/2);
    if (y%2 == 0)
        return temp*temp;
    else
        return x*temp*temp;
}

ll modulo( ll value, ll m)
{
    ll mod = value % m;
    if (value < 0) {
        mod += m;
    }
    return mod;
}

vector<ll> initializeDiffArray(vector<ll>& A)
{
    ll n = A.size();

    // We use one extra space because
    // update(l, r, x) updates D[r+1]
    vector<ll> D(n + 1);

    D[0] = A[0], D[n] = 0;
    for (ll i = 1; i < n; i++)
        D[i] = A[i] - A[i - 1];
    return D;
}

// Does range update
void update(vector<ll>& D, ll l, ll r, ll x)
{
    D[l] += x;
    D[r + 1] -= x;
}

// Prints updated Array
void getArray(vector<ll>& A, vector<ll>& D)
{
    for (ll i = 0; i < A.size(); i++) {
        if (i == 0)
            A[i] = D[i];

        // Note that A[0] or D[0] decides
        // values of rest of the elements.
        else
            A[i] = D[i] + A[i - 1];
    }
}