• POJ1579 Function Run Fun


    Function Run Fun

    Time Limit: 1000ms
    Memory Limit: 10000KB
    This problem will be judged on PKU. Original ID: 1579
    64-bit integer IO format: %lld      Java class name: Main
     
    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
    

    Source

     
    解题:一个寒假没写代码了啊。。。。。。。先找下感觉啊。。。
     
     1 /*
     2 @author: Lev
     3 @date:
     4 */
     5 #include <iostream>
     6 #include <cstdio>
     7 #include <cmath>
     8 #include <cstring>
     9 #include <string>
    10 #include <cstdlib>
    11 #include <algorithm>
    12 #include <map>
    13 #include <set>
    14 #include <queue>
    15 #include <climits>
    16 #include <deque>
    17 #include <sstream>
    18 #include <fstream>
    19 #include <bitset>
    20 #include <iomanip>
    21 #define LL long long
    22 #define PI acos(-1)
    23 using namespace std;
    24 int dp[30][30][30];
    25 int go(int a,int b,int c) {
    26     if(a <= 0 || b <= 0 || c <= 0) return 1;
    27     if(a > 20 || b > 20 || c > 20) return go(20,20,20);
    28     if(dp[a][b][c]) return dp[a][b][c];
    29     if(a < b && b < c) dp[a][b][c] = go(a,b,c-1)+go(a,b-1,c-1)-go(a,b-1,c);
    30     else dp[a][b][c] = go(a-1, b, c) + go(a-1, b-1, c) + go(a-1, b, c-1) - go(a-1, b-1, c-1);
    31     return dp[a][b][c];
    32 }
    33 int main() {
    34     memset(dp,0,sizeof(dp));
    35     int a,b,c;
    36     while(~scanf("%d %d %d",&a,&b,&c)) {
    37         if(a == -1 && b == -1 && c == -1) break;
    38         printf("w(%d, %d, %d) = %d
    ",a,b,c,go(a,b,c));
    39     }
    40     return 0;
    41 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4319217.html
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