一个递归算法求f(m,n)的问题.

设m.n均为自然数，m可表示为一些不超过n的自然数之和，f(m,n)为这种表示方式的数目。
例f(5,3)=5，有5种表示方式：3+2，3+1+1，2+2+1，2+1+1+1，1+1+1+1+1。递归算法求f(m,n).

 

 

f(m, n)
可以对照图表，不难观察出来，总共有这样几种情况了

 

(1). m < n

f(1, 2) = f(1, 3) = f(1, 4) = ... = f(1, 1)
f(2, 3) = f(2, 4) = f(2, 5) = ... = f(2, 2)
......
得出结论：
m < n : f(m, n) = f(m, m)
/*====================================================*/

 

 

(2). n = 1

 

f(1, 1) = f(2, 1) = f(3, 1) = ... = 1
得出结论：
n = 1 : f(m, n) = 1;
/*====================================================*/

 

 

(3). m = n

 

f(2, 2) = 2 = 1+1 = 1+f(2, 1)
f(3, 3) = 3 = 1+2 = 1+f(3, 2)
f(4, 4) = 4 = 1+3 = 1+f(4, 3)
......
m = n : f(m, n) = 1+f(m, m-1)
/*====================================================*/

 

(4). m > n

m-n = 1

f(3, 2) = 2 = f(3, 1) + 1 = 1+1
f(4, 3) = 4 = f(4, 2) + 1 = 3+1
f(5, 4) = 6 = f(5, 3) + 1 = 5+1
......
而f(m, n) = 1的只有f(m, 1),所以我们可以用f(1, 1)统一代替
于是有了，当m-n=1时
f(m, n) = f(m, n-1) + f(1, 1)
---------------------------

 

m-n = 2

 

f(4, 2) = 3 = f(4, 1) + 2 = 1+2
f(5, 3) = 5 = f(5, 2) + 2 = 3+2
f(6, 4) = 9 = f(6, 3) + 2 = 7+2
......
而f(m, n) = 2的只有f(2, 2)与f(3,2)
于是有了，当m-n = 2时
f(m, n) = f(m, n-1) + f(2, 2)
f(m, n) = f(m, n-1) + f(3, 2)
---------------------------

 

m-n = 3

f(5, 2) = 3  = f(5, 1) + 2 = 1+2
f(6, 3) = 7  = f(6, 2) + 3 = 4+3
f(7, 4) = 11 = f(7, 3) + 3 = 8+3
......
而f(m, n) = 2的有f(3, 2), f(2, 2)
而f(m, n) = 3的有f(3, 3), f(4, 2), f(5, 2)
于是有了，当m-n = 3时
m = 5时
f(m, n) = f(m, n-1) + f(3, 2) // ?
f(m, n) = f(m, n-1) + f(2, 2) // f((5-2),(5-2))
------------------
m > 5时
f(m, n) = f(m, n-1) + f(3, 3) // f((6-3),(6-3)) or f((7-4),(7-4))
f(m, n) = f(m, n-1) + f(4, 2) // ?
f(m, n) = f(m, n-1) + f(5, 2) // ?
------------------
所以可以推出来
当m > n时
f(m, n) = f(m, n-1) + f(m-n, m-n))

不过别忘记了，我们之前有对m=n的情况下，把所有的值增加1了，然后才去运算的。
所以我们还要针对m=n的做减1的动作。

if (m == n)
{
 nRet--;
}

 

附上代码：

#include "stdio.h"#define TEST_NUM 10#define SPACE_LINE ("----------------------------------------") /* 40 - */int func(int nM, int nN){ if ((1 == nM) || (1 == nN)) { return 1; } if (nM < nN) { return (nM, nM); } if (nM == nN) { return (1 + func(nM, nM-1)); } return (func(nM, nN-1) + func(nM-nN, nN));}int main(void){ int nM; int nN; int nRet; for (nM = 1; nM <= TEST_NUM; nM++) { for (nN = 1; nN <= nM; nN++) { nRet = func(nM, nN); if (nM == nN) { nRet--; } printf("nRet(%2d) = func(nM(%2d), nN(%2d))/n", nRet, nM, nN); } printf("%s/n", SPACE_LINE); } return 0;}