动态规划【 Function Run Fun】
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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.
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
#include <iostream>using namespace std;long long a,b,c,d;int dp[21][21][21];int w(int i,int j,int k){ if(i <= 0 || j <= 0 || k <= 0) return 1; return dp[i][j][k];}void W(){ int i,j,k; for(i = 1; i < 21; i++) for(j = 1; j < 21; j++) for(k = 1; k < 21; k++) { if(i < j && j < k) dp[i][j][k] = w(i,j,k-1) + w(i,j-1,k-1) - w(i,j-1,k); else dp[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 Ww(int i,int j,int k){ if(i <= 0 || j <= 0 || k <= 0) return 1; if(i > 20 || j > 20 || k > 20) return dp[20][20][20]; else return dp[i][j][k];}int main(){ W(); while(cin>>a>>b>>c) { if(a==-1&&b==-1&&c==-1) break; else d=Ww(a,b,c); cout<<"w("<<a<<", "<<b<<", "<<c<<") "<<"= "<<d<<endl; }return 0;}
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