1.
题目大意：
给一个长度为n的的数组a，n是1到1e5，ai是1到1e5，问a的所有排序的序列的逆序对之和，会有重复的数字出现
题目链接：https://ac.nowcoder.com/acm/contest/46597/E
typedef long long ll;
typedef long long LL;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef pair<double,double> pdd;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
#define endl '\n'
const int N = 1e5 + 7;
LL ksm(LL x, LL y) {
LL res = 1;
while (y) {
if (y & 1) {
res = 1ll * res * x % MOD;
}
x = 1ll * x * x % MOD;
y >>= 1;
}
return res;
}
LL inv(int x) {
return ksm(x, MOD - 2);
}
LL b[N], fac[N];
signed main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
int n;
std::cin >> n;
std::vector<LL>a(n + 1), s(n + 1), cnt(n + 1);
for (int i = 1; i <= n; ++i) {
std::cin >> a[i];
b[i] = a[i];
}
if (n == 1) {
std::cout << 0 << '\n';
return 0;
}
std::sort(b + 1, b + 1 + n);
int len = std::unique(b + 1, b + 1 + n) - b - 1;
for (int i = 1; i <= n; ++i) {
a[i] = std::lower_bound(b + 1, b + 1 + len, a[i]) - b;
++cnt[a[i]];
}
fac[0] = 1;
for (int i = 1; i <= n; ++i) {
fac[i] = 1ll * fac[i - 1] * i % MOD;
}
LL ans = 0;
for (int i = 1; i <= len; ++i) {
s[i] = s[i - 1] + cnt[i];
}
for (int i = 1; i <= len; ++i) {
ans = (ans + s[i - 1] * cnt[i]) % MOD;
}
std::cout << 1ll * ans * fac[n] % MOD * inv(2) % MOD << '\n';
return 0;
}