Problem:
Score : 100 points

Problem Statement

We have five variables x_1, x_2, x_3, x_4, and x_5.

The variable x_i was initially assigned a value of i.

Snuke chose one of these variables and assigned it 0.

You are given the values of the five variables after this assignment.

Find out which variable Snuke assigned 0.

Constraints

The values of x_1, x_2, x_3, x_4, and x_5 given as input are a possible outcome of the assignment by Snuke.

Input

Input is given from Standard Input in the following format:

x_1 x_2 x_3 x_4 x_5

Output

If the variable Snuke assigned 0 was x_i, print the integer i.

Sample Input 1

0 2 3 4 5

Sample Output 1

1

In this case, Snuke assigned 0 to x_1, so we should print 1.

Sample Input 2

1 2 0 4 5

Sample Output 2

3

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)


#include <bits/stdc++.h>

#define fast                          \
    ios_base::sync_with_stdio(false); \
    cin.tie(0);                       \
    cout.tie(0);

#define ll long long
#define ld long double

#define pb push_back
#define ins insert
#define er erase

#define f(i, a, b) for (int i = a; i < b; ++i)
#define rf(i, a, b) for (int i = a; i >= b; --i)
#define cmax(a, b) a = max(a, b)
#define cmin(a, b) a = min(a, b)
#define cadd(a, b) a = add(a, b)
#define csub(a, b) a = subt(a, b)
#define cmul(a, b) a = mult(a, b)
#define mem(a, b) memset(a, b, sizeof(a))

#define vi vector<int>
#define vst vector<string>
#define vpii vector<pii>

#define all(v) v.begin(), v.end()
#define lb lower_bound
#define ub upper_bound
#define mxe max_element
#define mne min_element

#define srt(v) sort(all(v))
#define rsrt(v) sort(v.rbegin(), v.rend())

#define pii pair<int, int>
#define ff first
#define ss second

#define bs bitset
#define bpc(a) __builtin_popcount(a)

#define mii map<int, int>
#define umii unordered_map<int, int>

#define si set<int>
#define usi unordered_set<int>
#define spii set<pii>

#define pqi priority_queue<int>
#define pqpii priority_queue<pii>

#define dbg(a, st, end)                   \
    f(i, st, end + 1) cerr << a[i] << sp; \
    cerr << nl;
#define pnt(a, st, end)                   \
    f(i, st, end + 1) cout << a[i] << sp; \
    cout << nl;
#define dbgc(a)          \
    for (auto y : a)     \
        cerr << y << sp; \
    cerr << nl;
#define dbgm(a)                            \
    for (auto y : a)                       \
        cerr << y.ff << "," << y.ss << sp; \
    cerr << nl;

#define trace(a) cerr << a << nl;
#define trace2(a, b) cerr << a << sp << b << nl;
#define trace3(a, b, c) cerr << a << sp << b << sp << c << nl;
#define trace4(a, b, c, d) cerr << a << sp << b << sp << c << sp << d << nl;

#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*1ll * (b)) / gcd(a, b)

#define nl "\n"
#define sp " "

using namespace std;
using namespace std::chrono;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define ok order_of_key
#define fo find_by_order

//ll md = 1000000007;
ld pi = 3.141592653;
ll md = 998244353;

ll pw(ll a, ll b)
{
    ll c = 1, m = a;
    while (b)
    {
        if (b & 1)
            c = (c * m);
        m = (m * m);
        b /= 2;
    }
    return c;
}

ll pwmd(ll a, ll b)
{
    ll c = 1, m = a;
    while (b)
    {
        if (b & 1)
            c = (c * m) % md;
        m = (m * m) % md;
        b /= 2;
    }
    return c;
}

ll modinv(ll n)
{
    return pwmd(n, md - 2);
}

ll my_ceil(ll a, ll b)
{
    if (a % b == 0)
        return a / b;
    return a / b + 1;
}

ll nc2(ll n)
{
    return (1ll * n * (n - 1)) / 2;
}

ll nsum(ll n)
{
    return (1ll * n * (n + 1)) / 2;
}

int my_log(ll n, int b)
{
    ll i = 1;
    int ans = 0;
    while (1)
    {
        if (i > n / b)
            break;
        i *= b;
        ans++;
    }
    return ans;
}

inline int nc()
{
    static char buf[100000], *p1 = buf, *p2 = buf;
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;
}
inline int read()
{
    int ret = 0;
    bool f = 0;
    char ch = nc();
    while (ch > '9' || ch < '0')
        f ^= ch == '-', ch = nc();
    while (ch <= '9' && ch >= '0')
        ret = ret * 10 + ch - '0', ch = nc();
    return f ? -ret : ret;
}

int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};

inline ll add(ll a, ll b) { return (((a >= md) ? a % md : a) + ((b >= md) ? b % md : b) + md) % md; }
inline ll subt(ll a, ll b) { return (a - b + md) % md; }
inline ll mult(ll a, ll b) { return (1ll * ((a >= md) ? a % md : a) * ((b >= md) ? b % md : b)) % md; }

inline int manhatten(pii p1, pii p2) { return abs(p1.ff - p2.ff) + abs(p1.ss - p2.ss); }

inline bool kthbit(ll n, int k) { return (n >> k) & 1; }
inline ll setkthbit(ll n, int k) { return n | (1ll << k); }
inline ll unsetkthbit(ll n, int k) { return n & ~(1ll << k); }
inline ll flipkthbit(ll n, int k) { return n ^ (1ll << k); }

//LET'S PLAY THE GAME!!

/*

CHECK FOR N=1, N=2, etc;
REMOVE DEBUG STATEMENTS
CHECK FOR DUPLICATES (SET -> MULTISET)
CHECK FOR OVERFLOW
CLEAR EVERYTHING BEFORE EVERY TESTCASE
DON'T MISTAKE ROOT FOR LEAF
DON'T ALWAYS IGNORE THE BRUTE FORCE SOLUTION!

*/

#define int long long //UNCOMMENT IF NEEDED

int32_t main()
{
    fast;

    int m, h, c, e, k;
    int x, q, l, r, y, n;

    string s, t, u;

    auto start = high_resolution_clock::now();

    int tcnt = 0;

    int tests = 1;
    //pre();

    //cin >> tests;

    while (tests--)
    {
        //cout << "Case #" << ++tcnt << ": ";
        int a[5];
        f(i, 0, 5) cin >> a[i];
        f(i, 0, 5) if (!a[i])
        {
            cout << i + 1;
            return 0;
        }
    }
    auto stop = high_resolution_clock::now();
    auto duration = duration_cast<microseconds>(stop - start);
    cerr << "Time taken: "
         << duration.count() / 1000000.0 << "seconds" << nl;
    return 0;
}

// AND THE GAME IS OVER !!