HDU 1331 Function Run Fun
来源:互联网 发布:网络奇葩女唯一的承诺 编辑:程序博客网 时间:2024/06/07 13:09
Function Run Fun
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2799 Accepted Submission(s): 1349
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.
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
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int N = 21 ;int dp[N][N][N] ;int w ( int a , int b , int c ){ if ( a <= 0 || b <= 0 || c <= 0 ) return 1 ; if ( a > 20 || b > 20 || c > 20 ) return w( 20 , 20 ,20 ); if ( dp [a][b][c]) return dp[a][b][c]; if ( a < b && b < c ) return ( dp[a][b][c] = w ( a , b ,c -1 )+ w(a,b-1 ,c-1) - w(a , b -1 , c) ); return ( dp[a][b][c] = w ( a-1 ,b,c) + w(a-1 ,b-1 ,c) + w( a - 1 , b , c -1 ) - w (a-1 , b-1 ,c-1)) ;}int main(){ int a , b ,c ; while ( cin >> a>>b >> c ){ if ( a == -1 && b == -1 &&c == -1 ) break; else { memset( dp , 0 , sizeof(dp)) ; printf("w(%d, %d, %d) = %d\n" , a,b,c, w(a,b,c) ); } } return 0;}
0 0
- hdu 1331 Function Run Fun
- HDU 1331 Function Run Fun
- HDU 1331 Function Run Fun
- HDU 1331 Function Run Fun
- HDU Function Run Fun
- Function Run Fun HDU
- hdu 1331 Function Run Fun(DP)
- hdu acm 1331 1579 Function Run Fun
- 【HDU 1331 Function Run Fun】+ 记忆搜索
- hdu 1579 Function Run Fun
- hdu 1579 Function Run Fun
- [递归理解/记忆化] HDU/HOJ 1331 Function Run Fun
- hdu 1331||1579 Function Run Fun(记忆化)
- hdu 1331 Function Run Fun(记忆化搜索)
- HDU--1331--Function Run Fun--记忆化搜索
- HDU 1331--Function Run Fun【水题】【打表】
- HDU 1331 Function Run Fun (基础记忆化搜索)
- HDU 1331 Function Run Fun(记忆化搜索)
- 深入理解linux内核读书笔记(第十章)
- nim manual(一)
- centos 安装jdk
- 前端的几个好用软件
- 从头开始写项目Makefile(三):变量的使用
- HDU 1331 Function Run Fun
- meta大全
- JVM中堆栈以及内存区域分配
- 队列(Queue) C 语言实现
- HDU 2993 MAX Average Problem (斜率优化)
- LeetCode之Word Break II
- hibernate 多对一单向映射配置文件的配置
- Java名词解释
- windwos 8三大版本的区别