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;
}