Saving Beans
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2648 Accepted Submission(s): 983
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
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/*
考慮加多一顆樹，這樣的話當加的樹放了k(0<=k<=m)個beans時，原本的n顆樹上放的beans數量之和就等於m-k（<=m）,滿足題目的要求 ，也降低了計算的難度。
則題目要求的是a1+a2+......an+an+1=m（0<=ai<=m,1<=i<=n+1) 式1
解有多少組。
考慮把問題轉換成，求a1+a2+......an+an+1=m+n+1（1<=ai<=m+1,1<=i<=n+1) 式2
解有多少組。
因為式1的每組解，對於每個ai，都加上1的話，就是式2的一組解。
對於式2的求解：
考慮有m+n+1個Beans排成一列，則它們中恰好有m+n個間隔，在m+n個間隔中選擇n個各插入一塊木板，則把這些Beans分成n+1部分，每部分的值對應到每個ai，就是式2的一組解。而在m+n個間隔中選擇n個，則是求組合數的問題了，p<=10^5且為質數，則可用Lucas定理求。
以下轉載於：
http://blog.csdn.net/tju_virus/article/details/7843248
題目相當於求n個數的和不超過m的方案數。
如果和恰好等於m，那麼就等價於方程x1+x2+...+xn = m的解的個數，利用插板法可以得到方案數為：
(m+1)*(m+2)...(m+n-1) = C(m+n-1,n-1) = C(m+n-1,m)
現在就需要求不大於m的，相當於對i = 0,1...,m對C(n+i-1,i)求和，根據公式C(n,k) = C(n-1,k)+C(n-1,k-1)得
C(n-1,0)+C(n,1)+...+C(n+m-1,m)
= C(n,0)+C(n,1)+C(n+1,2)+...+C(n+m-1,m)
= C(n+m,m)
現在就是要求C(n+m,m) % p,其中p是素數。
然後利用Lucas定理的模板就可以輕松的求得C(n+m,m) % p的值
簡單介紹一下Lucas定理：
Lucas定理是用來求 C(n,m) mod p的值,p是素數（從n取m組合，模上p）。 描述為: Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p) Lucas(x,0,p)=1;
A、B是非負整數，p是質數。AB寫成p進制：A=a[n]a[n-1]...a[0]，B=b[n]b[n-1]...b[0]。
則組合數C(A,B)與C(a[n],b[n])*C(a[n-1],b[n-1])*...*C(a[0],b[0]) modp同余
即：Lucas(n,m,p)=c(n%p,m%p)*Lucas(n/p,m/p,p)
簡單的理解就是：
以求解n! % p 為例，把n分段，每p個一段，每一段求得結果是一樣的。但是需要單獨處理每一段的末尾p,2p,...,把p提取出來，會發現剩下的數正好又是(n/p)! ，相當於
劃歸了一個子問題，這樣遞歸求解即可。
這個是單獨處理n！的情況，當然C(n,m)就是n！/(m! *（ｎ－ｍ）！），每一個階乘都用上面的方法處理的話，就是Lucas定理了
Lucas最大的數據處理能力是p在10^5左右。
而C(a,b) =a! / ( b! * (a-b)! ) mod p
其實就是求 ( a! / (a-b)!) * ( b! )^(p-2) mod p
(上面這一步變換是根據費馬小定理：假如p是質數，且a,p互質，那麼a的（p-1）次方除以p的余數恆為1,
那麼a和a^(p-2)互為乘法逆元，則（b / a) = (b * a^(p-2) ) mod p)
*/ #include
#include #include #include using namespace std; templateinline T read(T&x) { char c; while((c=getchar())<=32)if(c==EOF)return 0; bool ok=false; if(c=='-')ok=true,c=getchar(); for(x=0; c>32; c=getchar()) x=x*10+c-'0'; if(ok)x=-x; return 1; } template inline T read_(T&x,T&y) { return read(x)&&read(y); } template inline T read__(T&x,T&y,T&z) { return read(x)&&read(y)&&read(z); } template inline void write(T x) { if(x<0)putchar('-'),x=-x; if(x<10)putchar(x+'0'); else write(x/10),putchar(x%10+'0'); } templateinline void writeln(T x) { write(x); putchar('\n'); } //-------ZCC IO template------ const int maxn=11; const double inf=999999999; #define lson (rt<<1),L,M #define rson (rt<<1|1),M+1,R #define M ((L+R)>>1) #define For(i,t,n) for(int i=(t);i<(n);i++) typedef long long LL; typedef double DB; typedef pair P; #define bug printf("---\n"); //#define mod 10007 LL powmod(LL a,LL b,LL mod) { LL rec=1; while(b) { if(b&1)rec=(rec*a)%mod; a=(a*a)%mod; b>>=1; } return rec%mod; } LL C(LL a,LL b,LL mod) { LL ans=1,ca=1,cb=1; if(aa-b)b=a-b; for(int i=0;i
