HDU 5976 Detachment 【贪心】 (2016ACM/ICPC亚洲区大连站)

Detachment

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 570    Accepted Submission(s): 192

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:

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

4

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5976

题目大意：

　　给一个数N(N<=10⁹)，让你把它拆成若干各不相同的数Ai，ΣAi=N，要求ΠAi（累乘）最大。

题目思路：

　　【贪心】

　　首先肯定要把位数拆的尽量多，手写了20以内的拆法。

　　发现以2为首相的递增序列累乘最大，所以我的想法就是把N拆成2+3+...+x<=n，

　　先找到x，之后算一下n还多了多少，就把后面依次+1，变成2+3+...+y+(y+2)+(y+3)+...+(x+1)。

　　这时候它们的累乘是最大的。

　　（特殊情况是从2到x都加1之后还剩余1，这时候把最后一项再加1，变成3+4+...+x+(x+2)

  1 //
  2 //by coolxxx
  3 /*
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<string>
  7 #include<iomanip>
  8 #include<map>
  9 #include<stack>
 10 #include<queue>
 11 #include<set>
 12 #include<bitset>
 13 #include<memory.h>
 14 #include<time.h>
 15 #include<stdio.h>
 16 #include<stdlib.h>
 17 #include<string.h>
 18 #include<math.h>
 19 //#include<stdbool.h>
 20 #define min(a,b) ((a)<(b)?(a):(b))
 21 #define max(a,b) ((a)>(b)?(a):(b))
 22 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
 23 */
 24 #include<bits/stdc++.h>
 25 #pragma comment(linker,"/STACK:1024000000,1024000000")
 26 #define abs(a) ((a)>0?(a):(-(a)))
 27 #define lowbit(a) (a&(-a))
 28 #define sqr(a) ((a)*(a))
 29 #define mem(a,b) memset(a,b,sizeof(a))
 30 #define eps (1e-8)
 31 #define J 10000
 32 #define mod 1000000007
 33 #define MAX 0x7f7f7f7f
 34 #define PI 3.14159265358979323
 35 #define N 45004
 36 using namespace std;
 37 typedef long long LL;
 38 double anss;
 39 LL aans;
 40 int cas,cass;
 41 int n,m,lll,ans;
 42 LL a[N],ni[N];
 43 LL mi(LL x,LL y)
 44 {
 45     LL z=1;
 46     while(y)
 47     {
 48         if(y&1)z=(z*x)%mod;
 49         x=(x*x)%mod;
 50         y>>=1;
 51     }
 52     return z;
 53 }
 54 void init()
 55 {
 56     int i;
 57     a[1]=1;
 58     ni[1]=1;
 59     for(i=2;i<N;i++)
 60     {
 61         a[i]=(a[i-1]*i)%mod;
 62         ni[i]=(-(mod/i)*a[mod%i])%mod;
 63     }
 64 }
 65 int main()
 66 {
 67     #ifndef ONLINE_JUDGE
 68     freopen("1.txt","r",stdin);
 69 //    freopen("2.txt","w",stdout);
 70     #endif
 71     int i,j,k;
 72     int x,y,z;
 73     init();
 74     for(scanf("%d",&cass);cass;cass--)
 75 //    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
 76 //    while(~scanf("%s",s))
 77 //    while(~scanf("%d%d",&n,&m))
 78     {
 79         scanf("%d",&n);
 80         if(n<5)
 81         {
 82             printf("%d
",n);
 83             continue;
 84         }
 85         m=n+n+2;
 86         LL l,r,mid;
 87         l=2;r=45000;
 88         while(l<r)
 89         {
 90             mid=(l+r+1)>>1;
 91             if(mid*mid+mid<=m)l=mid;
 92             else r=mid-1;
 93         }
 94         m-=l*l+l;
 95         m/=2;
 96         if(m==l)
 97         {
 98             aans=a[l]*(l+2)%mod*mi(2,mod-2)%mod;
 99         }
100         else
101         {
102             x=l+1-m;
103             aans=a[l+1]*mi(x,mod-2)%mod;
104         }
105         printf("%lld
",aans);
106     }
107     return 0;
108 }
109 /*
110 //
111 
112 //
113 */

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