POJ 1579

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Description

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.

Output

Print the value for w(a,b,c) for each triple.

Sample Input

1 1 12 2 210 4 650 50 50-1 7 18-1 -1 -1

Sample Output

w(1, 1, 1) = 2w(2, 2, 2) = 4w(10, 4, 6) = 523w(50, 50, 50) = 1048576w(-1, 7, 18) = 1
题意：大家爱递归！。说实话我怎么爱递归。简单来讲。根据题目给的要求对一组数字进行递归。常规的dfs会超时，所以进行记忆化搜索。没啥子好说的。
#include<stdio.h>#include<string.h>int dp[100][100][100];int dfs(int x,int y,int z){ if(x<=0||y<=0||z<=0)                     //判断条件，根据题意来就行了，{    return 1;}if(dp[x][y][z]>0)                    //如果已经找到，。直接return；return dp[x][y][z];   if(x>20||y>20||z>20){return dfs(20,20,20);}else if(x<y&&y<z){  dp[x][y][z]=dfs(x,y,z-1)+dfs(x,y-1,z-1)-dfs(x,y-1,z);}else { dp[x][y][z]=dfs(x-1,y,z)+dfs(x-1,y-1,z)+dfs(x-1,y,z-1)-dfs(x-1,y-1,z-1);}return dp[x][y][z];            }int main(){int a,b,c;while(scanf("%d%d%d",&a,&b,&c)!=EOF){if(a==-1&&b==-1&&c==-1)break;int i,j,k;for(i=0;i<=21;i++)for(j=0;j<=21;j++)for(k=0;k<=21;k++)dp[i][j][k]=0;printf("w(%d, %d, %d) = %d\n",a,b,c,dfs(a,b,c));}return 0;}

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