poj 1579 Function Run Fun

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.

 

题意很好理解，不解释

数组还函数的名字不要起一样的，要大小写区分

分析，由于子问题过多，每次递归的时候就记录结果

#include<stdio.h>
#include<string.h>

const int MAXN=25;

int w[MAXN][MAXN][MAXN];

int W(int a,int b,int c)
{
    if(w[a][b][c]) return w[a][b][c];
    if(a<=0 || b<=0 || c<=0) return w[a][b][c]=1;
    if(a<b && b<c) return w[a][b][c]=W(a,b,c-1)+W(a,b-1,c-1)-W(a,b-1,c);
    else return w[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);
}

int main()
{
    int a,b,c;
    while(scanf("%d%d%d",&a,&b,&c))
    {
        memset(w,0,sizeof(w));
        if(a==-1 && b==-1 && c==-1) break;
        if(a<=0 || b<=0 || c<=0) printf("w(%d, %d, %d) = 1\n",a,b,c);
        else if(a>20 || b>20 || c>20) printf("w(%d, %d, %d) = %d\n",a,b,c,W(20,20,20));
        else     printf("w(%d, %d, %d) = %d\n",a,b,c,W(a,b,c));
    

    }
    return 0;
}

从别人那学来的

预处理

void init()
{
    for(a=0;a<=20;a++)
        for(b=0;b<=20;b++)
            for(c=0;c<=20;c++)
                m[a][b][c]=1;
    for(a=1;a<=20;a++)
        for(b=1;b<=20;b++)
            for(c=1;c<=20;c++){
                if(a < b && b < c)
                    m[a][b][c]=m[a][b][c-1]+m[a][b-1][c-1]-m[a][b-1][c];
                else
                    m[a][b][c]=m[a-1][b][c]+m[a-1][b-1][c]+m[a-1][b][c-1]-m[a-1][b-1][c-1];
            }
}