HDU1331 Function Run Fun

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3754    Accepted Submission(s): 1836

Problem 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
#include<iostream>#include<stdio.h>#include<cmath>#include<string>#include<string.h>#include<queue>#include<algorithm>using namespace std;int re[25][25][25];     //a,b,c∈（0,20）int i,j,k;int a,b,c;int f(int a,int b,int c){    if(a<=0||b<=0||c<=0) return 1;    if(a>20||b>20||c>20) return f(20,20,20);    if(re[a][b][c])return re[a][b][c];        //此if一定要在第三个位置，因为数组开的范围<25    if(a<b&&b<c) re[a][b][c]=f(a, b, c-1) + f(a, b-1, c-1) - f(a, b-1, c);    else re[a][b][c]=f(a-1, b, c) + f(a-1, b-1, c) + f(a-1, b, c-1) - f(a-1, b-1, c-1);    return re[a][b][c];}int main(){    memset(re,0,sizeof(re));    while(cin>>a>>b>>c)    {        if(a==-1&&b==-1&&c==-1)break;       printf("w(%d, %d, %d) = %d\n",a,b,c,f(a,b,c));    }    return 0;}