HDU 5976 Detachment 打表找规律

题目链接：

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

Detachment

Time Limit: 4000/2000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others) #### 问题描述 > 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,a2, … (x= a1+a2+…) assigned to different dimensions. And then, the multidimensional space has been established. Now there are two requirements for this space: > 1．Two different small line segments cannot be equal ( ai≠aj when i≠j). > 2．Make this multidimensional space size s as large as possible (s= 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) #### 输入 > 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 #### 输出 > Maximum s you can get modulo 10^9+7. Note that we wants to be greatest product before modulo 10^9+7. ####样例输入 > 1 > 4

样例输出

4

题意

给你一个数n，求把它拆分成若干个数的和，使得这些数的乘积最大。

题解

打表找了下规律：

2
阶段1(3~4)
2 3
阶段2(6~8)
2 3 4
阶段3(10~13)
2 3 4 5
...

处理下阶乘和阶乘的逆，二分找下n属于哪个阶段，然后把区间里面的阶乘拆两部分求就可以了（求x*(x+1)...y利用前缀和思想）。

代码

include

include

include

include

include

include

include

include

include

include

include

include

include

include

include

using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
//#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-6;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e5+10;
const int mod=1e9+7;

LL fac[maxn],invfac[maxn],inv[maxn];
void pre(){
fac[0]=fac[1]=1;
invfac[0]=invfac[1]=1;
inv[1]=1;
for(int i=2;i<maxn;i++){
fac[i]=fac[i-1]i%mod;
inv[i]=(mod-mod/i)inv[mod%i]%mod;
invfac[i]=invfac[i-1]*inv[i]%mod;
}
}

int n;

inline LL f(LL x){
return (x+1)*x/2-1;
}

int main() {
pre();
int tc;
scf("%d",&tc);
while(tc--){
scf("%d",&n);
if(n<=4){
prf("%d\n",n);
continue;
}
int l=2,r=100000;
while(l+1<r){
int mid=l+(r-l)/2;
if(f(mid)<=n) l=mid;
else r=mid;
}

int len=l-1;

LL ans=1;
if(f(l+1)-n==1){
ans=fac[3+len-2]invfac[2]%mod;
ans=ansfac[l+2]%modinvfac[l+1]%mod;
}else{
int d=n-f(l);
ans=fac[l+1]invfac[l+1-d]%mod;
ans=ans*fac[l-d]%mod;
}

prf("%lld\n",ans);

}
return 0;
}

//end-----------------------------------------------------------------------