Problem:
Score : 100 points

Problem Statement

There are three airports A, B and C, and flights between each pair of airports in both directions.

A one-way flight between airports A and B takes P hours, a one-way flight between airports B and C takes Q hours, and a one-way flight between airports C and A takes R hours.

Consider a route where we start at one of the airports, fly to another airport and then fly to the other airport.

What is the minimum possible sum of the flight times?

Constraints

1 \leq P,Q,R \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

P Q R

Output

Print the minimum possible sum of the flight times.

Sample Input 1

1 3 4

Sample Output 1

4

The sum of the flight times in the route A \rightarrow B \rightarrow C: 1 + 3 = 4 hours

The sum of the flight times in the route A \rightarrow C \rightarrow C: 4 + 3 = 7 hours

The sum of the flight times in the route B \rightarrow A \rightarrow C: 1 + 4 = 5 hours

The sum of the flight times in the route B \rightarrow C \rightarrow A: 3 + 4 = 7 hours

The sum of the flight times in the route C \rightarrow A \rightarrow B: 4 + 1 = 5 hours

The sum of the flight times in the route C \rightarrow B \rightarrow A: 3 + 1 = 4 hours

The minimum of these is 4 hours.

Sample Input 2

3 2 3

Sample Output 2

5

Solution in C++14 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional>
#include <string>
#include <queue>
#include <cmath>
#include <iomanip>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <random>
#include <unordered_map>
#include <unordered_set>
#define ll long long
#define fr(i,k,N) for(ll i=k; i<N;i++)
#define rep(i,N) for(ll i=0; i<N;i++)
#define ALL(v) v.begin(),v.end()
#define mp make_pair
using namespace std;
template <class X> void pr(X test){cout<<test<<endl;}
template <class X> void prr(X test){for(auto it:test){cout<<it<<endl;}}
template <class X> void prrr(X test){int f=0;for(auto it:test){cout<<(f++!=0?" ":"")<<it;}cout<<endl;}
template<typename X>
istream&operator>>(istream&i,vector<X>&v){for(X&x:v)i>>x;return i;}

class UF { // union find from https://github.com/kartikkukreja/blog-codes (MIT)
    // [usage]
    // auto uf = new UF(n);
    // uf->merge(l, r); // 0 <= l, r <= n-1
    // ...
public:
    int *id; // inner id
    int cnt; // number of all vartices
    int *sz; // sz[i] means size of i's set
    UF(int N)   {
        cnt = N;
        id = new int[N];
        sz = new int[N];
        for(int i=0; i<N; i++) {
            id[i] = i;
            sz[i] = 1;
        }
    }
    ~UF() {
        delete [] id;
        delete [] sz;
    }
    int find(int p) {
        int root = p;
        while (root != id[root]) root = id[root];
        while (p != root) {
            int newp = id[p];
            id[p] = root;
            p = newp;
        }
        return root;
    }
    void merge(int x, int y) {
        int i = find(x);
        int j = find(y);
        if (i == j) return;
        
        if (sz[i] < sz[j]) {
            id[i] = j;
            sz[j] += sz[i];
        } else {
            id[j] = i;
            sz[i] += sz[j];
        }
        cnt--;
    }
    bool connected(int x, int y) {
        return find(x) == find(y);
    }
    int count() {
        return cnt;
    }
};

// return sorted 約数s 1 <= x <= n
vector<ll> divv(ll n) {
  vector<ll> v;
  for(ll i = 1; i*i <= n; i++) {
    if(n%i == 0) {
      v.push_back(i);
      if(i*i != n) v.push_back(n/i);
    }
  }
  sort(ALL(v));
  return v;
}

bool chk(string &s, ll a,ll b){
    fr(i,a - 1, b - 1){
        auto j = i +1;
        if(s[i] == '#' && s[j]== '#') {
            return false;
        }
    }
    return true;
}

int main() {
    ll n,x,m;
    cin>>n>>x>>m;
    pr(min(n+x,min(x+m,m+n)));

    
    return 0;
}