HDU-5976 Detachment

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【HDU-5976 Detachment】

题意： 给定一个数字分成若干段， 要求每一段的长度不相等， 要求这些段的乘积最大。
分析： 这是一个标准的Maximum Product Over Partitions Into Distinct Parts 问题，打表找规律， 参见论文

#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;

typedef long long ll;
const ll mod = 1e9 + 7;
const int MAXN = 1e5 + 10;
ll tr[MAXN];
ll fa[MAXN];
void get_tr () {
    for (int i = 1; i < MAXN; i++) 
        tr[i] = (long long)(i + 1)*(long long)i/2;
}

void get_fa() {
    fa[0] = 1;
    for (int i = 1; i < MAXN; i++) {
        fa[i] = ((long long)i*fa[i-1])%mod%mod;
    }
}

long long extend_gcd (ll a, ll b, ll &x, ll &y) {
    if (a == 0 && b == 0) return -1;
    if (b == 0) {
        x = 1; y = 0; return a;
    }
    long long d = extend_gcd(b, a%b, y, x);
    y -= a/b*x;
    return d;
}

ll inv (ll a, ll m) {
    long long x, y;
    long long d = extend_gcd (a, m, x, y);
    if (d == 1) return (x%m + m)%m;
    else return -1;
}

int main () {
    int T;
    long long n;
    get_fa();
    get_tr();

    scanf ("%d", &T);
    /*
    for (int i = 1; i < 100; i++) {
        cout << tr[i] << endl;
    }
    */
    while (T--) {
        scanf ("%lld", &n);
        
        if (n == 1){
            printf("1\n");
            continue;
        }
        ll l = 0;
        int m = lower_bound (tr + 1, tr + MAXN - 1, n) - tr;
        if (tr[m] == n) l = 0;
        else {
            m = m - 1;
            l = n - tr[m];
        }
        /*
        for (int i = 1; i < MAXN; i++) {
            if (n < tr[i]) {
                m = i - 1;
                break;
            }
        }
        */


        
        ll t = fa[m];
        ll p = (ll)m;
        if (l >= 0 && l <= p - 2) {
            printf ("%lld", ((t*(p + 1))%mod*inv(p - l, mod))%mod%mod);
        } else if (l == m - 1) {
            printf ("%lld", (t*(p + 2))%mod*inv(2, mod)%mod)%mod;
        } else if (l == m) {
            printf ("%lld", t*(p+1)%mod%mod);
        }
        printf("\n");
    }
    return 0;
}