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  • POJ1579 Function Run Fun

    Function Run Fun
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 13517   Accepted: 7040

    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: 


    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

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 
     5 using namespace std;
     6 
     7 int dp[21][21][21];
     8     
     9 int calw(int a, int b, int c)
    10 {
    11     if(a <= 0 || b <= 0 || c <= 0)
    12     {
    13         //[a][b][c] = 1;   //注意这里不能直接写,以为a、b、c可能超出数组范围
    14         return 1;
    15     }
    16     if(a > 20 || b > 20 || c > 20)
    17     {
    18         dp[20][20][20] = calw(20, 20, 20);
    19         return dp[20][20][20];
    20     }
    21     if(dp[a][b][c])
    22         return dp[a][b][c];
    23     if(a < b && b < c)
    24     {
    25         dp[a][b][c] = calw(a, b, c - 1) + calw(a, b - 1, c - 1) - calw(a, b-1, c);
    26         return dp[a][b][c];
    27     }
    28     dp[a][b][c] = calw(a-1, b, c) + calw(a - 1, b - 1, c) + calw(a - 1, b, c - 1) - calw(a - 1, b - 1, c - 1);
    29     return dp[a][b][c];
    30 }
    31 
    32 int main()
    33 {
    34     int a, b, c;
    35     while(cin >> a >> b >> c && a != -1 || b != -1 || c != -1)
    36     {
    37         printf("w(%d, %d, %d) = %d\n",a, b, c, calw(a, b, c));
    38     }
    39     return 0;
    40 }
    功不成,身已退
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  • 原文地址:https://www.cnblogs.com/dongsheng/p/2779367.html
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