动态规划（2）Function Run Fun

                                                                                                                                                                                                                                                                                                               Function Run Fun

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 14127 Accepted: 7344

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.

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

Source

Pacific Northwest 1999

题很简单，就是继续强调不能只想着用递归来做，而是用最基础的DP思想来做，不多说了，下面是代码：

/*#include<stdio.h>long long fun(int a,int b,int c){if(a<=0 || b<=0 || c<=0) return 1;else if(a>20 || b>20 || c>20) return fun(20,20,20);else if(a<b && b<c) return fun(a, b, c-1) + fun(a, b-1, c-1) - fun(a, b-1, c);else return fun(a-1, b, c) + fun(a-1, b-1, c) + fun(a-1, b, c-1) - fun(a-1, b-1, c-1) ;};int main(){int a,b,c,i,j,k;long long ans;while(scanf("%d %d %d",&a,&b,&c)!= EOF){if(a==-1 && b==-1 && c==-1)   break;else {ans = fun(a,b,c);printf("%lld\n",ans);}}return 0;}*///上述方法会超时。。。。。。。//用一个三维数组直接把所有的解都算出来存起来，就行了。注意初始化为1#include<stdio.h>long long w[22][22][22];int i,j,k;int a,b,c;void t(){ for(i = 0;i<=20;i++)for(j = 0;j<=20;j++)for(k = 0;k<=20;k++)w[i][j][k] = 1;//初始化    for(i = 1;i<=20;i++)for(j = 1;j<=20;j++)for(k = 1;k<=20;k++){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];//计算出所有值        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];}};int main(){t();while(scanf("%d %d %d",&a,&b,&c)!= EOF){if(a==-1 && b==-1 && c==-1)   break;else if(a<=0 || b<=0 || c<=0) printf("w(%d, %d, %d) = 1\n",a,b,c);else if(a>20 || b>20 || c>20)  printf("w(%d, %d, %d) = %lld\n",a,b,c,w[20][20][20]);else{printf("w(%d, %d, %d) = %lld\n",a,b,c,w[a][b][c]);}}return 0;}