HDU1331 Function Run Fun
2016-05-27 22:50
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Function Run Fun
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3754 Accepted Submission(s): 1836
Problem 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 1
2 2 2
10 4 6
50 50 50
-1 7 18
-1 -1 -1
Sample Output
w(1, 1, 1) = 2
w(2, 2, 2) = 4
w(10, 4, 6) = 523
w(50, 50, 50) = 1048576
w(-1, 7, 18) = 1
#include<iostream> #include<stdio.h> #include<cmath> #include<string> #include<string.h> #include<queue> #include<algorithm> using namespace std; int re[25][25][25]; //a,b,c∈(0,20) int i,j,k; int a,b,c; int f(int a,int b,int c) { if(a<=0||b<=0||c<=0) return 1; if(a>20||b>20||c>20) return f(20,20,20); if(re[a][b][c])return re[a][b][c]; //此if一定要在第三个位置,因为数组开的范围<25 if(a<b&&b<c) re[a][b][c]=f(a, b, c-1) + f(a, b-1, c-1) - f(a, b-1, c); else re[a][b][c]=f(a-1, b, c) + f(a-1, b-1, c) + f(a-1, b, c-1) - f(a-1, b-1, c-1); return re[a][b][c]; } int main() { memset(re,0,sizeof(re)); while(cin>>a>>b>>c) { if(a==-1&&b==-1&&c==-1)break; printf("w(%d, %d, %d) = %d\n",a,b,c,f(a,b,c)); } return 0; }
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