poj1579解题报告

题目大姨：给出一个递归法则，求出给定变量a,b,c经过该递归后的值..法则如下

 

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)

 

 

Sample Input

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

Sample Output

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

刚开始我直接按照题目要求写的递归函数，发现当a,b,c较大时不能在1s内得出答案，考虑到递归过程中有很多重复的递归，于是想到开个三维数组来记录w(a,b,c)的值。。编完发现运行结果不对，找了半天错误最后发现递归法则34最后一个符号是减。。，改过来0ms水果了。。

 

 

#include<iostream>using namespace std;long int dw[21][21][21];long int w(int a,int b,int c){int sum=0;if(a<=0 || b<=0 || c<=0)return 1;if(a>20||b>20||c>20){if(dw[20][20][20]==-1)dw[20][20][20]=w(20,20,20);return dw[20][20][20];}if(a<b&&b<c){if(c-1>0){if(dw[a][b][c-1]==-1)dw[a][b][c-1]=w(a,b,c-1);sum+=dw[a][b][c-1];}elsesum++;if(b-1>0 && c-1>0){if(dw[a][b-1][c-1]==-1)dw[a][b-1][c-1]=w(a,b-1,c-1);sum+=dw[a][b-1][c-1];}elsesum++;if(b-1>0){if(dw[a][b-1][c]==-1)dw[a][b-1][c]=w(a,b-1,c);sum-=dw[a][b-1][c];}elsesum--;return sum;}if(a-1>0){if(dw[a-1][b][c]==-1)dw[a-1][b][c]=w(a-1,b,c);sum+=dw[a-1][b][c];}elsesum++;if(a-1>0&&b-1>0){if(dw[a-1][b-1][c]==-1)dw[a-1][b-1][c]=w(a-1,b-1,c);sum+=dw[a-1][b-1][c];}elsesum++;if(a-1>0 &&c-1>0){if(dw[a-1][b][c-1]==-1)dw[a-1][b][c-1]=w(a-1,b,c-1);sum+=dw[a-1][b][c-1];}elsesum++;if(a-1>0&&b-1>0&&c-1>0){if(dw[a-1][b-1][c-1]==-1)dw[a-1][b-1][c-1]=w(a-1,b-1,c-1);sum-=dw[a-1][b-1][c-1];}elsesum--;returnsum; }int main(){memset(dw,-1,sizeof(dw));int a,b,c;while(cin>>a>>b>>c &&(a+1||b+1||c+1))cout<<"w("<<a<<", "<<b<<", "<<c<<") = "<<w(a,b,c)<<endl;return 0;}