1579 Function Run Fun

Function Run Fun

Time Limit:

1000ms

Memory limit:

10000kB

题目描述

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.

输入

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.

输出

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

样例输入

1 1 1
2 2 2
10 4 6
50 50 50
-1 7 18
-1 -1 -1

样例输出

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

#include<stdio.h>

int opt[100][100][100];
int w(int a,int b,int c)
{
    if(a <= 0 || b <= 0 || c <= 0) return 1;
    if(opt[a][b][c]!=-1) return opt[a][b][c];
    if(a > 20 || b > 20||c > 20){
           opt[a][b][c]=w(20, 20, 20);
               return w(20, 20, 20);
               }
    if(a < b && b < c){
         opt[a][b][c]=w(a, b, c-1)+w(a, b-1, c-1)-w(a, b-1, c);
         return w(a, b, c-1)+w(a, b-1, c-1)-w(a, b-1, c);
         }
         opt[a][b][c]=w(a-1,b,c)+w(a-1,b-1,c)+w(a-1,b,c-1)-w(a-1,b-1,c-1);
     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;
    for(a=0;a<100;a++) for(b=0;b<100;b++) for(c=0;c<100;c++) opt[a][b][c]=-1;
    while(scanf("%d%d%d",&a,&b,&c),!(a==-1 && b==-1 && c==-1))
    {
       printf("w(%d, %d, %d) = %d/n",a,b,c,w(a,b,c));
    }
    return 0;
}