程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> hdu3037 Saving Beans（Lucas定理+費馬小定理or擴展歐幾裡德算發）

hdu3037 Saving Beans（Lucas定理+費馬小定理or擴展歐幾裡德算發）

編輯：關於C++

Saving Beans

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3221 Accepted Submission(s): 1234

Problem Description Although winter is far away, squirrels have to work day and night to save beans. They need plenty of food to get through those long cold days. After some time the squirrel family thinks that they have to solve a problem. They suppose that they will save beans in n different trees. However, since the food is not sufficient nowadays, they will get no more than m beans. They want to know that how many ways there are to save no more than m beans (they are the same) in n trees.

Now they turn to you for help, you should give them the answer. The result may be extremely huge; you should output the result modulo p, because squirrels can’t recognize large numbers.
Input The first line contains one integer T, means the number of cases.

Then followed T lines, each line contains three integers n, m, p, means that squirrels will save no more than m same beans in n different trees, 1 <= n, m <= 1000000000, 1 < p < 100000 and p is guaranteed to be a prime.
Output You should output the answer modulo p.
Sample Input

2
1 2 5
2 1 5

Sample Output

3
3
Hint
Hint

For sample 1, squirrels will put no more than 2 beans in one tree. Since trees are different, we can label them as 1, 2 … and so on. 
The 3 ways are: put no beans, put 1 bean in tree 1 and put 2 beans in tree 1. For sample 2, the 3 ways are:
 put no beans, put 1 bean in tree 1 and put 1 bean in tree 2.

Source 2009 Multi-University Training Contest 13 - Host by HIT
題意：將不大於m顆種子存放在n顆樹中，問有多少種存法（注意是不大於，並不是等於）。

分析：首先是不大於m顆種子，我沒可以認為少於m的那些種子存放在了第n+1顆樹上，這樣的話，問題就轉化成了將m顆種子存放在n+1顆樹上的方案數。ok這個是組合數學裡面的公式，亦即插板法，也就是X1+X2+X3+……+Xn+1 = m；so，答案是C(n+m,m)。然後就是基礎的費馬小定理求逆元（當然也可以用擴展歐幾裡德算法求逆元，然而我才剛開始學數論，並不會 - -），Lucas定理求解了。

費馬小定理不懂的點這裡；

擴展歐幾裡德算法可以點這裡；

Lucas定理不懂的點這裡。

#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

ll f[100010];

ll init(ll p)//求階乘，用於計算組合數
{
    f[0]=1;
    for (int i=1; i<=p ;i++)
        f[i] = f[i-1]*i%p;
}

ll mod_pow(int a, int n, int p)//這裡求的是a^n，也就是(f[bb]*f[aa-bb]%p)^(p-2)
{
    ll aa = a,re = 1;
    while (n)
    {
        if (n&1) re=(re*aa)%p;
        aa=(aa*aa)%p;
        n>>=1;
    }
    return re;
}

ll Lucas(ll a, ll b, ll p)//Lucas定理
{
    ll re=1;
    while (a && b)
    {
        ll aa=a%p,bb=b%p;
        if (aa < bb) return 0;
        re =re*f[aa]*mod_pow(f[bb]*f[aa-bb]%p, p-2, p)%p;//費馬小定理求逆元
        a /= p;
        b /= p;
    }
    return re;
}

int main ()
{
    int T;
    ll n,m,p;
    cin>>T;
    while (T--)
    {
        cin>>n>>m>>p;
        init(p);
        cout<