CSU-ACM2017暑期训练3-递推与递归 D-Function Run Fun POJ-1579

题目：

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. 

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

 

             

             题目大意：给你一个递归式，但是这个递归很慢，要想输出结果需要你进行改进。

                思路：将递归改成记忆化搜索或者递推式都可以。我这里用的递推式。注意坑点，最后几个判断输入的合法性的写法。

代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int w[22][22][22];int init(){    for(int a=0;a<=20;a++)    {        for(int b=0;b<=20;b++)        {            for(int c=0;c<=20;c++)            {                if(a==0||b==0||c==0)                    w[a][b][c]=1;                else if(a<b&&b<c)                    w[a][b][c]=w[a][b][c-1]+w[a][b-1][c-1]-w[a][b-1][c];                else                    w[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(){    memset(w,0,sizeof(w));    int x,y,z;    init();    while(scanf("%d%d%d",&x,&y,&z)!=EOF)    {        if(x==-1&&y==-1&&z==-1)//这里常喜欢写成x+y+z!=-1但是本题不行            break;        if(x<=0||y<=0||z<=0)//根据题意首先要判断是否小于等于0            printf("w(%d, %d, %d) = %d\n",x,y,z,w[0][0][0]);        else if(x>20||y>20||z>20)//这里不能打等于号            printf("w(%d, %d, %d) = %d\n",x,y,z,w[20][20][20]);        else            printf("w(%d, %d, %d) = %d\n",x,y,z,w[x][y][z]);    }    return 0;}