poj_1579 && hdoj_1331-白红宇

poj_1579 && hdoj_1331

阅读量：3529 次

发布时间：2019-05-20

本文共 2427 字，大约阅读时间需要 8 分钟。

Function Run Fun

Time Limit:1000MS		Memory Limit:10000K
Total Submissions:13554		Accepted:7056

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 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

之前用记忆化搜索写过。

三个for循环，从w[0][0][0]递推到w[20[20][20]

注意题目的优先计算顺序。先是考虑是否会小于0，然后才考虑是否会大于20，要不然就会WA。

//#include
   
    //#include
    
     //#include
     
      //using namespace std;//int w[100][100][100];////int DFS(int a,int b,int c)//{//	if(a<=0 || b<=0 || c<=0)//	{//		return 1;//	}//	if(a>20 || b>20 || c>20)//	{//		return DFS(20,20,20);//	}//	if(w[a][b][c]!=-1)//	{//		return w[a][b][c];//	}//	if(a
      >i>>j>>k)//	{//		if(i==-1&&j==-1&&k==-1)//			break;//		printf("w(%d, %d, %d) = %d\n",i,j,k,DFS(i,j,k));//	}//}#include
       
        #include
        
         #include
         
          using namespace std;int w[25][25][25];int main(){ freopen("in.txt","r",stdin); int i, j, k, a, b, c; memset(w, 0, sizeof(w)); for(i=0; i<=20; i++) { for(j=0; j<=20; j++) { for(k=0; k<=20; k++) { if(i == 0 || j == 0 || k == 0) { w[i][j][k] = 1; } else if(i < j && j < k) { w[i][j][k] = w[i][j][k-1] + w[i][j-1][k-1] - w[i][j-1][k]; } else { w[i][j][k] = w[i-1][j][k] + w[i-1][j-1][k] + w[i-1][j][k-1] - w[i-1][j-1][k-1]; } } } } while(cin >> a >> b >> c) { if(a == -1 && b == -1 && c == -1) break; if(a <= 0 || b <= 0 || c <= 0) printf("w(%d, %d, %d) = %d\n",a,b,c,1); else if(a > 20 || b > 20 || c > 20) printf("w(%d, %d, %d) = %d\n",a,b,c,w[20][20][20]); else printf("w(%d, %d, %d) = %d\n",a,b,c,w[a][b][c]); } return 0;}

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