思路统计最大值出现的次数，和最小值出现的次数。虽然是每次都是MAX-MIN，我们先求MAX的和，然后再求MIN的和，做差。 这次代码写的真的很漂亮 题目地址：
代码语言：javascript复制#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;
}
//---------------https://lunatic.blog.csdn.net/-------------------//
#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 100005
#define fil(a, n) rep(0, n, i) read(a[i])
ll power(ll a, ll b, ll p)
{
ll ans = 1 % p;
for (; b; b >>= 1)
{
if (b & 1)
ans = ans * a % p;
a = a * a % p;
}
return ans;
}
long long mm[500000];
void init(ll n, ll k)
{
mm[1] = 1;
for (ll i = 2; i <= n; i )
{
mm[i] = ((mm[i - 1] * (k i - 1)) % MOD * power(i - 1, MOD - 2, MOD)) % MOD;
//cout<<mm[i]<<endl;
}
}
ll a[N];
int main()
{
int n, k;
read(n), read(k);
fil(a, n);
sort(a, a n);
init(n - k 2, k - 1);
long long sum = 0;
rrep(n - 1, k - 2, i)
sum = (sum (mm[i - k 2] * (a[i] - a[n - i - 1]) % MOD) % MOD) % MOD;
wl((sum MOD) % MOD);
P;
}
