Function Run Fun
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 16503 | Accepted: 8514 |
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 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
巨水。。
一開始wa了一组数据 （0,21,21）
#include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <string> #include <cctype> #include <vector> #include <cstdio> #include <cmath> #include <deque> #include <stack> #include <map> #include <set> #define ll long long #define maxn 116 #define pp pair<int,int> #define INF 0x3f3f3f3f #define max(x,y) ( ((x) > (y)) ? (x) : (y) ) #define min(x,y) ( ((x) > (y)) ?(y) : (x) ) using namespace std; int n,m,dp[22][22][22]; int dfs(int x,int y,int z) { if(x<=0||y<=0||z<=0) return 1; if(x>20||y>20||z>20) return dfs(20,20,20); if(dp[x][y][z]!=INF) return dp[x][y][z]; dp[x][y][z]=0; if(x<y&&y<z) dp[x][y][z]+=(dfs(x,y,z-1)+dfs(x,y-1,z-1)-dfs(x,y-1,z)); else dp[x][y][z]+=(dfs(x-1,y,z)+dfs(x-1,y-1,z)+dfs(x-1,y,z-1)-dfs(x-1,y-1,z-1)); return dp[x][y][z]; } int main() { int a,b,c; while(scanf("%d%d%d",&a,&b,&c)) { if(a==-1&&b==-1&&c==-1)break; memset(dp,INF,sizeof(dp)); printf("w(%d, %d, %d) = %d ",a,b,c,dfs(a,b,c)); } return 0; }
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