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 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

 

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;}