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很容易想到最优策略是2*3*4一直乘到前缀和小于n的最靠后的位置,设这个位置为p
当segma(2,p)==n时明显答案就是mul(2,p)
但是当segma(2,p) != n时会有一个余项,设为need,need=n-segma(2,p),显然现在项数不会再增加了,我们只能把need按照一定的策略分配给前面的元素才能使结果更大,按照直觉我们应该将need平摊给前面所有的元素,但是要注意need有可能会分配多1个,这时要给最后一项分配2
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5;
const int INF = 1e9;
const int mod = 1e9 + 7;
typedef long long ll;
ll sum[maxn], mul[maxn], inv[maxn];
int cnt;
ll mpow(ll x, int y) {
ll ans = 1;
while (y) {
if (y & 1) ans = ans * x % mod;
x = x * x % mod;
y >>= 1;
}
return ans;
}
void init() {
sum[0] = sum[1] = 0;
mul[0] = mul[1] = 1;
for (int i = 2; i < maxn; i++) {
sum[i] = sum[i-1]+i;
mul[i] = mul[i-1]*i % mod;
if (sum[i] >= INF) break;
cnt = i;
}
inv[cnt] = mpow(mul[cnt], mod-2);
inv[0] = inv[1] = 0;
for (int i = cnt-1; i >= 2; i--) inv[i] = inv[i+1]*(i+1)%mod;
}
int main() {
init();
int T; scanf("%d", &T);
while (T--) {
int n; scanf("%d", &n);
if (n <= 4) {
printf("%d\n", n); continue;
}
int p = upper_bound(sum+1, sum+1+cnt, n)-sum-1;
ll need = n - sum[p], ans = 0;
if (need == 0) {
ans = mul[p];
}
else if (need == p) {
ans = mul[p]*inv[2]%mod*(p+2)%mod;
}
else {
ans = mul[p-need]*mul[p+1]%mod*inv[p-need+1]%mod;
}
printf("%lld\n", ans%mod);
}
return 0;
}