HDU 1331--Function Run Fun【水题】【打表】

Function Run Fun

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2572    Accepted Submission(s): 1250

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 1
2 2 2
10 4 6
50 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
   
   
   
   

 

不要想太复杂，数据比较小，打表就A了

#include <cstdio>
#include <cstring>
using namespace std;

int map[21][21][21];

void dabiao(){
    int i,j,k;
    for(i=0;i<=20;++i)
        for(j=0;j<=20;++j)
            for(k=0;k<=20;++k){
                if(i<=0 ||j<=0 ||k<=0)
                    map[i][j][k]=1;
                else if(i<j && j<k)
                    map[i][j][k]=map[i][j][k-1]+map[i][j-1][k-1]-
                    map[i][j-1][k];
                else
                    map[i][j][k]=map[i-1][j][k]+map[i-1][j-1][k]+
                    map[i-1][j][k-1]-map[i-1][j-1][k-1];
            }
    return ;
}

int main (){
    int a,b,c;
    dabiao();
    while(scanf("%d%d%d",&a,&b,&c)){
        if(a==-1 && b==-1 && c==-1)
            break;
        printf("w(%d, %d, %d) = ",a,b,c);
        if(a<=0 || b<= 0 || c<=0)
            printf("1\n");
        else if(a>20 || b>20 || c>20)
            printf("%d\n",map[20][20][20]);
        else
            printf("%d\n",map[a][b][c]);
    }
        return 0;
}