哈理工OJ 1029 Function Run Fun（记忆思想）

Function Run Fun
Time Limit: 1000 MS Memory Limit: 65536 K
Total Submit: 339(159 users) Total Accepted: 168(146 users) Rating: Special Judge: No
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

题目本身并不是太难，就是简单的递推，但难的地方就是你不知道把已经知道的结果储存起来，所以呢。在这个题中，我学会课记忆的思想，假如你已经算出了某个式子的结果，那么把他的结果储存在一个静态的数组里，下次再出现相同的问题的时候，直接返回答案就好了，省去了很多时间。
下面是AC代码：

#include<cstdio> #include<cstring> #include<algorithm> #define LL long long using namespace std;LL mem[21][21][21]={0}; LL fun(LL a,LL b,LL c) {if(a<=0||b<=0||c<=0){return 1;}if(a>20||b>20||c>20){return 1048576;}if(mem[a][b][c]){return mem[a][b][c];}if(a<b&&b<c){return mem[a][b][c]=fun(a,b,c-1)+fun(a,b-1,c-1)-fun(a,b-1,c);}else{return mem[a][b][c]=fun(a-1,b,c)+fun(a-1,b-1,c)+fun(a-1,b,c-1)-fun(a-1,b-1,c-1);} }int main() {LL a,b,c;while(~scanf("%lld%lld%lld",&a,&b,&c)){if(a==-1&&b==-1&&c==-1){break;}LL re=fun(a,b,c);printf("w(%lld, %lld, %lld) = %lld\n",a,b,c,re);}return 0; }

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