poj

Function Run Fun

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 18338		Accepted: 9372

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

Pacific Northwest 1999

这道题没有给数据的范围，我没有考虑dp，然后就当做数学题来做，就发现里一组规律，如果x<=y<=z，那么答案一定是 1<<x。但是仍然TEL。百度后才知道数据的范围，dp就好了。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double Pi = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = 10000 + 100;
int dp[100][100][100];
LL solve(LL a, LL b, LL c) {
    if(a <= 0 || b <= 0 || c <= 0)  return 1;
    if(a > 20 || b > 20 || c > 20)  return solve(20, 20, 20);
    if(dp[a][b][c]) return dp[a][b][c];
    if(a <= b && b <= c)  dp[a][b][c] = 1<<a;
    else  dp[a][b][c] = solve(a - 1, b, c) + solve(a-1, b-1, c) + solve(a-1, b, c-1) - solve(a-1, b-1, c-1);
    return dp[a][b][c];
}
int main() {
    LL a, b, c;
    while (scanf("%lld%lld%lld", &a, &b, &c) != EOF) {
        if (a == -1 && b == -1 && c == -1) break;
        printf("w(%lld, %lld, %lld) = %lld
", a, b, c, solve(a, b, c));
    }
    return 0;
}