poj - 1579 Function Run Fun 【dp】

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\n", a, b, c, solve(a, b, c));
}
return 0;
}