1626. Function Run Fun

简单，一次性过，先将20以内的数计算出来，然后判断即可

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. For example:

1 1 12 2 210 4 650 50 50-1 7 18-1 -1 -1

Output

Print the value for w(a,b,c) for each triple, like this:

w(1, 1, 1) = 2w(2, 2, 2) = 4w(10, 4, 6) = 523w(50, 50, 50) = 1048576w(-1, 7, 18) = 1

#include<iostream>using namespace std;int main(){int a,b,c,w[21][21][21],n;for(int i=0;i<=20;i++){for(int j=0;j<=20;j++){for(int k=0;k<=20;k++){if(!i||!j||!k)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];elsew[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(1){cin>>a>>b>>c;if(a==-1&&b==-1&&c==-1)break;if(a<=0||b<=0||c<=0)n=1;else if(a>20||b>20||c>20)n=w[20][20][20];elsen=w[a][b][c];cout<<"w("<<a<<", "<<b<<", "<<c<<") = "<<n<<endl;}return 0;}