We all love recursion! Don't we? Consider a three-parameter recursive function w(a, b, c): if a <= 0 or b <= 0 or c <= 0, then w(a, b, c) returns: 1 if a > 20 or b > 20 or c > 20, then w(a, b, c) returns: w(20, 20, 20) if a < b and b < c, then w(a, b, c) returns: w(a, b, c-1) + w(a, b-1, c-1) - w(a, b-1, c) otherwise it returns: w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1) This is an easy function to implement. The problem is, if implemented directly, for moderate values of a, b and c (for example, a = 15, b = 15, c = 15), the program takes hours to run because of the massive recursion.
Input The input for your program will be a series of integer triples, one per line, until the end-of-file flag of -1 -1 -1. Using the above technique, you are to calculate w(a, b, c) efficiently and print the result. For example: 1 1 1 2 2 2 10 4 6 50 50 50 -1 7 18 -1 -1 -1
Output Print the value for w(a,b,c) for each triple, like this: w(1, 1, 1) = 2 w(2, 2, 2) = 4 w(10, 4, 6) = 523 w(50, 50, 50) = 1048576
w(-1, 7, 18) = 1
题目大意：递归在进行很多次的时候耗时非常巨大，需要想办法避免
分析：这是一道递归的题目，但正如题目所说，直接递归时间会很长。题目中三个整数的范围都不大，可使用记忆化搜索，也就是打表的方法。
建立一个三维数组把所有的结果存进去。
#include <stdio.h> int f[21][21][21];//存储递归的中间结果 int w(int a,int b,int c) { if(a<=0||b<=0||c<=0) return 1; if(a>20||b>20||c>20) return w(20,20,20); if(f[a][b][c]>0) return f[a][b][c]; if(a<b&&b<c) return w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c); else return w(a-1,b,c)+w(a-1,b-1,c)+w(a-1,b,c-1)-w(a-1,b-1,c-1); } int main() { int a,b,c; //打表 int i,j,k; for(i=0;i<21;i++) for(j=0;j<21;j++) for(k=0;k<21;k++) f[i][j][k]=w(i,j,k); while(scanf("%d%d%d",&a,&b,&c)!=EOF) { if(a==-1&&b==-1&&c==-1) break; printf("w(%d, %d, %d) =",a,b,c); printf(" %d\n",w(a,b,c)); } return 0; }