Function Run Fun(递归转递推)

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1579

Problem：

Function Run Fun

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2223 Accepted Submission(s): 1170



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

Source

冬练三九之一



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#include<iostream>

#include<algorithm>

using namespace std;

int num[22][22][22];

int main()

{

int i,j,k,a,b,c;

for(i=0;i<21;i++)

{

for(j=0;j<21;j++)

{

for(k=0;k<21;k++)

{

if(!i||!j||!k)

num[i][j][k]=1;

else if(i<j&&j<k)

num[i][j][k]=num[i][j][k-1]+num[i][j-1][k-1]-num[i][j-1][k];

else

num[i][j][k]=num[i-1][j][k]+num[i-1][j-1][k]+num[i-1][j][k-1]-num[i-1][j-1][k-1];

}

}

}

while(scanf("%d%d%d",&a,&b,&c)==3)

{

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,num[20][20][20]);

else

printf("w(%d, %d, %d) = %d\n",a,b,c,num[a][b][c]);

}

return 0;

}