2016 ACM/ICPC 大连区域赛 F—Detachment【贪心+逆元】

本文深入探讨了一道经典的算法题目，即如何将一个整数n拆分成若干个不相同整数之和，使得这些整数的乘积达到最大。文章详细分析了贪心算法的应用，并提供了具体的代码实现，帮助读者理解并解决类似问题。

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http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1006&cid=736

Detachment

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 346 Accepted Submission(s): 97

Problem Description

In a highly developed alien society, the habitats are almost infinite dimensional space.
In the history of this planet,there is an old puzzle.
You have a line segment with x units’ length representing one dimension.The line segment can be split into a number of small line segments: a1,a2a1+a2
ai≠aj
a1∗a2*...).Note that it allows to keep one dimension.That's to say, the number of ai can be only one.
Now can you solve this question and find the maximum size of the space?(For the final number is too large,your answer will be modulo 10^9+7)

Input

The first line is an integer T,meaning the number of test cases.
Then T lines follow. Each line contains one integer x.
1≤T≤10^6, 1≤x≤10^9

Output

Maximum s you can get modulo 10^9+7. Note that we wants to be greatest product before modulo 10^9+7.

Sample Input

1 4

Sample Output

题意：

给你一个数n(n<=1e9)，让你把n拆成若干个不相同的数的和，且这些数的积是所有拆分方法中最大的。输出这些数的最大积对1e9+7取模。

分析：

要取乘积的最大，要把数尽量分成多个【贪心】,例如2+3+4+。。。。+ans,如果和为n，那就输出ans的阶乘；

如果比n大1，输出3*4*....*(ans-1)*(ans+1)

其他情况差为x输出2*....*(x-1)*(x+1)*......*ans

代码：

#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007
using namespace std;
ll a[100055];
ll n,ans;
bool ok(ll mid)
{
    return (mid+1)*mid/2-1>=n;
}
ll power(ll aa,ll  n)   //a的n次方mod
{
    ll ans=1;
    aa=aa%mod;
    while (n)
    {
        if(n&1)
        ans=(ans*aa)%mod;
        n>>=1;
        aa=(aa*aa)%mod;
    }
    return ans;
}
int main()
{
    ll t,i,j,l,r,pos,mid;
    ll tt;
    a[0]=1;
    for(i=1;i<=100005;i++)
        a[i]=(a[i-1]*i)%mod;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld",&n);
        l=1;
        r=n;
        if(n<=4)
        {
            printf("%lld\n",n);
            continue;
        }
        while(l<=r)
        {
            mid=(l+r)>>1;
            if(ok(mid))
            {
                r=mid-1;
                ans=mid;
            }
            else
                l=mid+1;
        }
        pos=ans;
        ans=ans*(ans+1)/2-1-n;
        if(ans==0)
        {
            printf("%lld\n",a[pos]);
            continue;
        }
        if(ans==1)
        {
            tt=a[pos-1]*(pos+1)%mod;
            tt=tt*power(2,mod-2)%mod;
            printf("%lld\n",tt);
            continue;
        }
        else
        {
            tt=a[pos];
            tt=tt*power(ans,mod-2)%mod;
            printf("%lld\n",tt);
        }
    }
}