POJ 1579 Function Run Fun
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Function Run Fun
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 19515 Accepted: 9866
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
Source
Pacific Northwest 1999
题目比较简单,但是虽然题目给了你递归的表达式,但是我们不能直接使用,不然的话,肯定超时,
我的做法相对简单,直接,用for循环打表,O(n^3)。但是比较坑的地方在于,要完全按照他递归的写法来循环
#include<cstdio>#include<algorithm>#include<set>#include<iostream>#include<cmath>#include<cstring>using namespace std;int f[25][25][25];int main (){ int a,b,c; memset(f,0,sizeof(0)); for(int i=0; i<21; i++) for(int j=0; j<21; j++) for (int k=0; k<21; k++) { if(!i||!j||!k) f[i][j][k]=1; else if(i<j&&j<k) f[i][j][k]=f[i][j][k-1]+f[i][j-1][k-1]-f[i][j-1][k]; else f[i][j][k]=f[i-1][j][k]+f[i-1][j-1][k]+f[i-1][j][k-1]-f[i-1][j-1][k-1]; } while(~scanf("%d%d%d",&a,&b,&c)) { if(a==-1&&b==-1&&c==-1) break; int d=a,e=b,h=c; if(a<=0||b<=0||c<=0) a=0,b=0,c=0;//这里的条件判断不能写反,题目是先判断是否小于0再判断是否大于20的 else if(a>20||b>20||c>20) a=20,b=20,c=20; printf("w(%d, %d, %d) = %d\n",d,e,h,f[a][b][c]); }}
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