Codeforces Round #509 (Div. 2) A. Heist 贪心

There was an electronic store heist last night.

All keyboards which were in the store yesterday were numbered in ascending order from some integer number x. For example, if x=4 and there were 3 keyboards in the store, then the devices had indices 4, 5 and 6, and if x=10 and there were 7 of them then the keyboards had indices 10, 11, 12, 13, 14, 15 and 16.

After the heist, only n keyboards remain, and they have indices a1,a2,…,an. Calculate the minimum possible number of keyboards that have been stolen. The staff remember neither x nor the number of keyboards in the store before the heist.

Input The first line contains single integer n (1≤n≤1000) — the number of keyboards in the store that remained after the heist.

The second line contains n distinct integers a1,a2,…,an (1≤ai≤109) — the indices of the remaining keyboards. The integers ai are given in arbitrary order and are pairwise distinct.

Output Print the minimum possible number of keyboards that have been stolen if the staff remember neither x nor the number of keyboards in the store before the heist.

Examples inputCopy 4 10 13 12 8 outputCopy 2 inputCopy 5 7 5 6 4 8 outputCopy 0 Note In the first example, if x=8 then minimum number of stolen keyboards is equal to 2. The keyboards with indices 9 and 11 were stolen during the heist.

In the second example, if x=4 then nothing was stolen during the heist.

#include <bits/stdc++.h>
using namespace std;
template <typename t>
void read(t &x)
{
    char ch = getchar();
    x = 0;
    t f = 1;
    while (ch < '0' || ch > '9')
        f = (ch == '-' ? -1 : f), ch = getchar();
    while (ch >= '0' && ch <= '9')
        x = x * 10 + ch - '0', ch = getchar();
    x *= f;
}

#define wi(n) printf("%d ", n)
#define wl(n) printf("%lld ", n)
#define rep(m, n, i) for (int i = m; i < n; ++i)
#define rrep(m, n, i) for (int i = m; i > n; --i)
#define P puts(" ")
typedef long long ll;
#define MOD 1000000007
#define mp(a, b) make_pair(a, b)
#define N 100050
#define fil(a, n) rep(0, n, i) read(a[i])
//---------------https://lunatic.blog.csdn.net/-------------------//
long long a[N];
int main()
{
    int n;
    read(n);
    rep(0, n, i) read(a[i]);
    sort(a, a + n);
    long long ans = 0;
    reo(1, n, i) ans += a[i] - a[i - 1] - 1;
    wl(ans),P;
    return 0;
}