Partition
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 954 Accepted Submission(s): 545
Problem Description
How many ways can the numbers 1 to 15 be added together to make 15? The technical term for what you are asking is the "number of partition" which is often called P(n). A partition of n is a collection of positive integers (not necessarily
distinct) whose sum equals n.
Now, I will give you a number n, and please tell me P(n) mod 1000000007.
Now, I will give you a number n, and please tell me P(n) mod 1000000007.
Input
The first line contains a number T(1 ≤ T ≤ 100), which is the number of the case number. The next T lines, each line contains a number n(1 ≤ n ≤ 105) you need to consider.
Output
For each n, output P(n) in a single line.
Sample Input
4 5 11 15 19
Sample Output
7 56 176 490
Source
题意:问一个数n能被拆分成多少种方法
首先你要知道母函数+然后你要知道五边形数定理;
设第n个五边形数为,那么
,即序列为:1,
5, 12, 22, 35, 51, 70, ...
相应图形例如以下:
设五边形数的生成函数为。那么有:
以上是五边形数的情况。以下是关于五边形数定理的内容:
五边形数定理是一个由欧拉发现的数学定理。描写叙述欧拉函数展开式的特性。欧拉函数的展开式例如以下:
欧拉函数展开后,有些次方项被消去。仅仅留下次方项为1, 2, 5, 7, 12, ...的项次,留下来的次方恰为广义五边形数。
五边形数和切割函数的关系
欧拉函数的倒数是切割函数的母函数。亦即:
当中
为k的切割函数。
上式配合五边形数定理,有:
在 n>0 时,等式右側的系数均为0,比較等式二側的系数,可得
因此可得到切割函数p(n)的递归式:
比如n=10时,有:
所以,通过上面递归式,我们能够非常高速地计算n的整数划分方案数p(n)了。
详见维基百科:https://zh.wikipedia.org/wiki/%E4%BA%94%E8%A7%92%E6%95%B0#.E5.BB.A3.E7.BE.A9.E4.BA.94.E9.82.8A.E5.BD.A2.E6.95.B8 或 https://zh.wikipedia.org/wiki/%E4%BA%94%E9%82%8A%E5%BD%A2%E6%95%B8%E5%AE%9A%E7%90%86
转载请注明出处:http://blog.csdn.net/u010579068
详见代码:
#include<iostream> #include<cstdio> #define NN 100005 #define LL __int64 #define mod 1000000007 using namespace std; LL wu[NN],pa[NN]; void init() { pa[0]=1; pa[1]=1; pa[2]=2; pa[3]=3; LL ca=0; for(LL i=1;i<=100000/2;i++) { wu[ca++]=i*(3*i-1)/2; wu[ca++]=i*(3*i+1)/2; if(wu[ca-1]>100000) break; } for(LL i=4;i<=100000;i++) { pa[i]=(pa[i-1]+pa[i-2])%mod; ca=1; while(wu[2*ca]<=i) { if(ca&1) { pa[i]=(pa[i]-pa[i-wu[2*ca]])%mod; pa[i]=(pa[i]%mod+mod)%mod; if(wu[2*ca+1]<=i) pa[i]=(pa[i]-pa[i-wu[2*ca+1]])%mod; pa[i]=(pa[i]%mod+mod)%mod; } else { pa[i]=(pa[i]+pa[i-wu[2*ca]])%mod; pa[i]=(pa[i]%mod+mod)%mod; if(wu[2*ca+1]<=i) pa[i]=(pa[i]+pa[i-wu[2*ca+1]])%mod; pa[i]=(pa[i]%mod+mod)%mod; } ca++; } } } int main() { int T,n; init(); scanf("%d",&T); while(T--) { scanf("%d",&n); printf("%I64d\n",pa[n]); } return 0; }