zoj 1168 || poj 1579 Function Run Fun

递归 = =。。

 

题目的意思是，给你递归式，算。。。

 

开始没仔细看题，做到后来，瞟见一句话

 

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.

 

= =。好吧。怪我。没好好看题。本来想找规律的，找到一点规律不过不够用，然后就想着，打表吧。

 

一会一看 = =。。好做啊，每个数字只跟前面的数字有关系，好吧，打表吧 = =。。

 

超过20全按20算了，多好，果断打表过了。如果把这个数组存一下，再弄个新程序，直接打表，估计会更快~~

 

#include <cstdio>#include <cstdlib>#include <iostream>#include <string>using namespace std;int num[30][30][30];int w(int a,int b,int c){if( a <= 0 || b <= 0 || c <= 0 )return 1;if( a < b && b < c )return num[a][b][c-1] + num[a][b-1][c-1] - num[a][b-1][c];return num[a-1][b][c] + num[a-1][b-1][c] + num[a-1][b][c-1] - num[a-1][b-1][c-1];}int main(){int a,b,c;for(a=0; a<=20; a++)for(b=0; b<=20; b++)for(c=0; c<=20; c++)num[a][b][c] = w(a,b,c) ;while( scanf("%d%d%d",&a,&b,&c) != EOF ){if( a == b && b == c && c == -1 )break;if( a <= 0 || b <= 0 || c <= 0 ){printf("w(%d, %d, %d) = 1/n",a,b,c);continue;}if( a > 20 || b > 20 || c > 20 ){printf("w(%d, %d, %d) = %d/n",a,b,c,num[20][20][20]);continue;}printf("w(%d, %d, %d) = %d/n",a,b,c,num[a][b][c]);}return 0;}