Function Run Fun
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Time Limit: 1000MS |
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Memory Limit: 10000K |
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Total Submissions: 14943 |
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Accepted: 7738 |
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
Source
解题思路：
1、 按照题目说的写出程序，但运算结果会超时，所以要用一个三维数组来保存。
程序代码：
#include<stdio.h>
#include<string.h>
int d[22][22][22]; //用数组来保存之，这样就不会重复的调用函数了，可以节省很多时间
int w(int a,int b,int c)
{
if(a<=0||b<=0||c<=0) //小于 0 则返回。
return 1;
else if(a>20||b>20||c>20)
return d[20][20][20]=w(20,20,20); // 数组下标一定要是 20 不能是 abc 否则会溢出
if(d[a][b][c])
return d[a][b][c]; // 如果存在值，直接返回即可，不用再调用函数，可以节省时间
else if(a<b&&b<c)
return d[a][b][c]=w(a,b,c-1)+w(a,b-1,c-1)-w(a,b-1,c);
else
return d[a][b][c]=w(a-1,b,c)+w(a-1,b-1,c)+w(a-1,b,c-1)-w(a-1,b-1,c-1);
}
int main()
{
int a,b,c;
memset(d,0,sizeof(d));
while(scanf("%d %d %d",&a,&b,&c)&&(a!=-1||b!=-1||c!=-1))
{
printf("w(%d, %d, %d) = %d ",a,b,c,w(a,b,c));
}
return 0;
}