传送门

题意：

给你一个x=a1+a2+a3+...+an

让你求一个s=a1*a2*a3*...*an最大。(ai!=aj 当且仅当i!=j)

ans%1e9+7

思路：

经过分析得到要使s最大，那么a1*a2*a3*..一定要是连续的一串数的乘积，比如9就是 2*3*4  14就是2*3*4*5 如果这个数刚好在

9-14中间怎么办呢，经过一阵观察，可发现x-9剩下的可以在2*3*4的2 3 4里动手脚，比如10 我还剩了1，那么我是选择给4 +1还是

选择给2 +1呢，显然我给4 +1得到的结果是+2*3   +6 ，给2 +1的话会导致3重复，那么按照题意，就只能给4 +1 那么11怎么办

在10的基础上 2 3 5 那么我可以给3 +1 依次这样，那么 我如果是13怎么办呢

在 12的基础上 3 4 5 我再+1只能给5 +1 得到 3 4 6 ，那么14的话就是 2 3 4 5来的答案更大。

于是就得到了以下代码：

//china no.1
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef pair<long long int,long long int> ii;
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e5+10;
const int maxx=600005;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

LL fac[maxn];
LL a[maxn];
void ex_gcd(LL a, LL b, LL &x, LL &y, LL &d)
{
    if (!b) {d = a, x = 1, y = 0;}
    else{
        ex_gcd(b, a % b, y, x, d);
        y -= x * (a / b);
    }
}
LL inv2(LL t, LL p)
{//如果不存在，返回-1
    LL d, x, y;
    ex_gcd(t, p, x, y, d);
    return d == 1 ? (x % p + p) % p : -1;
}

void init()
{
    fac[0]=1;
    for(int i=1;i<=maxn;i++)
    {
        fac[i]=fac[i-1]*i;
        fac[i]%=mod;
    }
    for(int i=1;i<=maxn;i++)
    {
        a[i]=a[i-1]+i+1;
    }
}

int t,x;
int main()
{
    //freopen( "in.txt" , "r" , stdin );
    //freopen( "1.txt" , "w" , stdout );
    init();
    s_1(t);
    W(t--)
    {
        s_1(x);
        if(x==1)
        {
            puts("1");
            continue;
        }
        int pos=lower_bound(a+1,a+maxn+1,x)-a;
        if(a[pos]==x)
            print(fac[pos+1]);
        else
        {
            x-=a[pos-1];
            LL ans=1;
            if(x<=pos-1)
            {
                ans*=fac[pos-x];
                ans=ans*fac[pos+1]%mod*inv2(fac[pos+1-x],mod);
            }
            else
            {
                ans*=fac[pos+2];
                ans=ans*fac[pos]%mod*inv2(fac[pos+1]*fac[2]%mod,mod);
             }
            ans=(ans+mod)%mod;
            print(ans);
        }
    }
}