ACM学习历程—SNNUOJ 1116 A Simple Problem（递推 && 逆元 && 组合数学 && 快速幂）（2015陕西省大学生程序设计竞赛K题）

Description

Assuming a finite – radius “ball” which is on an N dimension is cut with a “knife” of N-1 dimension. How many pieces will the “ball” be cut into most?
However, it’s impossible to understand the following statement without any explanation.
Let me illustrate in detail.
When N = 1, the “ball” will degenerate into a line of finite length and the “knife” will degenerate into a point. And obviously, the line will be cut into d+1 pieces with d cutting times.
When N = 2, the “ball” will degenerate into a circle of finite radius and the “knife” will degenerate into a line. Likewise, the circle will be cut into (d^2+d+2)/2 pieces with d cutting times.
When N = 3, the “ball” will degenerate into a ball on a 3-dimension space and the “knife” will degenerate into a plane.
When N = 4, the “ball” will degenerate into a ball on a 4-dimension space and the “knife” will degenerate into a cube on a 3 dimension.
And so on.
So the problem is asked once again: Assuming a finite-radius “ball” which is on an N dimension is cut with a “knife” of N-1 dimension. How many “pieces” will the “ball” be cut into most?

Input

The first line of the input gives the number of test cases T. T test cases follow. T is about 300.
For each test case, there will be one line, which contains two integers N, d(1 <= N <= 10^5, 1 <= d <= 10^6).

Output

For each test case, output one line containing “Case #x: y”, where x is the test case number (standing from 1) and y is the maximum number of “pieces” modulo 10^9+7.

Sample Input

3

3 3

3 5

4 5

Sample Output

Case #1: 8

Case #2: 26

Case #3: 31

HINT

Please use %lld when using long long

 

题目大意就是n维空间切d刀，最多能分成几个部分。

基本上通过推倒三位的大概就能很快推出整个的递推式。

设p(n, d)表示n维空间切d刀。

假设已经切了d-1刀，最后一刀自然切得越多越好。于是最后一刀如果和所有d-1到切的话自然是最好。但是可以逆过来看，相当于d-1到切最后一刀这个n-1维空间。

于是p(n, d) = p(n, d-1) + p(n-1, d-1)

然而这个式子虽然出来了，但是根据n和d的范围打表是不可能的。也不能直接暴力递推求解，自然考虑到可能要直接求表达式。

然而，表达式我求了好久没求出来，不过看了最后表达式后，大概能有以下思路来求通项：

 

首先有以下事实：

1：手写打表的话：

 

d->	0	1	2	3	4	5
n
1	1	2	3	4	5	6
2	1	2	4	7	11	16
3	1	2	4	8	15	26
4	1	2	4	8	16	31
5	1	2	4	8	16	32
6	1	2	4	8	16	32

会发现当n >= d时，通项是2^d，其实稍微考虑一下确实如此。因为第一列都是1，自然第二列从第二项开始都是2，同理往后从对角线往后都是乘2，自然是2^d。

 

2：设p(n, d)的差数列为a(n, d)的话，

自然a(n, d) = p(n, d) – p(n-1, d)

由原式得p(n-1, d) = p(n-1, d-1) + p(n-2, d-1)

三式式消去p得

a(n, d) = a(n, d-1) + a(n-1, d-1)

说明p的差数列也是满足这个递推式，同理p的任意k阶差数列都满足这个式子。

然而让这些差数列最后通项不同的因素自然应该是前几项导致的

 

有了上面两个结论，于是只用求n < d的情况，可以从下面两个角度考虑

1：利用组合数式子：C(n, m) = C(n-1, m) + C(n-1, m-1)，其中C(n, m)表示从n个中取m个。

由于这个式子和题目递推式非常形似。 于是猜测C(n, m)为p的某一阶差数列。根据前几列和前几行的计算，C(n, m)为p的第一阶差数列。于是p(n, d) = sum(C(d, i)) (0 <= i <= n)

 

2：根据第一个结论：列出第一阶的差数列

d->	0	1	2	3	4	5
n
1		1	2	3	4	5
2	0	0	1	3	6	10
3	0	0	0	1	4	10
4	0	0	0	0	1	5
5	0	0	0	0	0	1
6	0	0	0	0	0	0

基本上可以找规律，发现第一阶差数列是C(n, m)。

 

然后就是求C(d, i)的和了，由于d很大，考虑C(d, i) = A(d, i) / i!，然后就是求分子和分母在模10^9+7的情况下的商了。自然需要考虑到逆元。

这里对于逆元的处理可以预处理打表，经测试直接在线求exgcd逆元会T掉。

这里预处理用了网上的一个神奇的递推式，还有一种是我大连海事一个同学的做法。

 

代码（神奇式子）：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#define LL long long
#define N 1000000007

using namespace std;

//快速幂
//O(logn)
LL quickPower(LL x, int n)
{
    x %= N;
    LL a = 1;
    while (n)
    {
        a *= n&1 ? x : 1;
        a %= N;
        n >>= 1;
        x = (x*x) % N;
    }
    return a;
}

LL c[100005], a[100005], inv[100005];
int n, d;

void init()
{
    //***预处理所有ｉ在质数MOD下的逆元
    inv[1] = 1;
    for (int i = 2; i <= 100000; i++)
        inv[i] = inv[N%i]*(N-N/i) % N;

    a[0] = 1;
    for (int i = 1; i <= 100000; ++i)
        a[i] = (inv[i]*a[i-1]) % N;
}

void work()
{
    if (n >= d)
    {
        printf("%lld\n", quickPower(2, d));
        return;
    }
    LL now = d, ans = 0;
    c[0] = 1;
    for (int i = 1; i <= n; ++i)
    {
        c[i] = (now*c[i-1]) % N;
        now--;
    }
    for (int i = 0; i <= n; ++i)
    {
        ans += c[i]*a[i];
        ans %= N;
    }
    printf("%lld\n", ans);
}

int main()
{
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    init();
    int T;
    scanf("%d", &T);
    for (int times = 1; times <= T; ++times)
    {
        printf("Case #%d: ", times);
        scanf("%d%d", &n, &d);
        work();
    }
    return 0;
}

代码二(exgcd):

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#define FOR(i,x,y)  for(int i = x;i < y;i ++)
#define IFOR(i,x,y) for(int i = x;i > y;i --)
#define ll  long long
#define N   111111
#define D   1111111
#define MOD 1000000007

using namespace std;

ll c[N],mu[N];
ll n,d;

ll quickpow(ll a,ll n,ll m){
    ll ans=1;
    while(n){
        if(n&1) ans = (ans*a)%m;
        a = (a*a)%m;
        n>>=1;
    }
    return ans;
}

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

ll inv(ll a,ll n){
    ll d,x,y;
    ex_gcd(a,n,d,x,y);
    return d == 1 ? (x+n)%n : -1;
}

void init(){
    FOR(i,1,N){
        mu[i] = inv(i,MOD);
    }
}

void C(){
    c[0] = 1;
    FOR(i,1,n+1){
        ll tem = (d+1-i)*mu[i]%MOD;
        c[i] = (tem*c[i-1]) % MOD;
    }
}

ll solve(){
    ll res = 0;
    FOR(i,0,n+1){
        res += c[i];
        res %= MOD;
    }
    return res;
}

int main()
{
    //freopen("test.in","r",stdin);
    int t,tCase = 0;
    scanf("%d",&t);
    init();
    while(t--){
        printf("Case #%d: ",++tCase);
        scanf("%lld%lld",&n,&d);
        ll ans = 0;
        if(n >= d){
            ans = quickpow(2,d,MOD);
        }
        else{
            C();
            ans = solve();
        }
        printf("%lld\n",ans);
    }
    return 0;
}