題目：

Problem 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

這道題貌似以前在哪兒做過。。。
題目意思很簡單，就是讓你輸入a, b, c的值，然后根據(jù)上面的遞歸規(guī)則求出最終的結(jié)果。
可能很多人一開始的時候看見上面的遞歸函數(shù)就認(rèn)為這道題很簡單：直接按照它的要求寫當(dāng)a, b, c滿足什么條件時就執(zhí)行對應(yīng)的遞歸式，但是這種情況對小的數(shù)據(jù)還可以，但是當(dāng)a, b, c取稍微大一點(diǎn)兒的值就慘了。比如a=15, b=15, c=15. 但我們可以發(fā)現(xiàn)，只要把它遞歸的過程畫出一小部分，我們就會發(fā)現(xiàn)其實(shí)這個過程中某些情況被遞歸了很多次，浪費(fèi)了不少的時間。但是只要我們中途當(dāng)?shù)谝淮芜f歸到這種情況時能把這個值存儲下來，之后的遞歸再遇到這種情況時就直接用這個值，無需再把這種情況遞歸下去。這樣我們就能節(jié)省許多時間。這就是記憶化搜索的基本思想吧。
而這道題實(shí)際上只要輸入的a, b, c有一個小于等于0，結(jié)果就為1；如果a, b, c中其中有一個值大于20，結(jié)果就是當(dāng)a=20,b=20,c=20時候的值。其他情況就采用記憶化搜索就可以了。

參考代碼：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 20+5;

int a, b, c;
LL dp[N][N][N];

void init() {
    memset(dp, -1, sizeof(dp));
}

LL solve(int a, int b, int c) {
    if (a <= 0 || b <= 0 || c <= 0) return 1;
    if (dp[a][b][c] >= 0) return dp[a][b][c];//記憶化搜索(如果搜索的值已經(jīng)存在, 則無需再次遞歸搜索)
    //else if (a <= 0 || b <= 0 || c <= 0) return 1;
    else if (a > 20 || b > 20 || c > 20) dp[a][b][c] = solve(20, 20, 20);
    else if (a < b && b < c) dp[a][b][c] = solve(a, b, c-1) + solve(a, b-1, c-1) - solve(a, b-1, c);
    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() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    
    init();
    dp[0][0][0] = 1;

    while (cin >> a >> b >> c) {
        if (a == -1 && b == -1 && c == -1) break;
    
        LL ans;
        if (!(a <= 0 || b <= 0 || c <= 0) && (a > 20 || b > 20 || c > 20)) {
            ans = solve(20, 20, 20);
        }
        else ans = solve(a, b, c);
        cout << "w(" << a << ", " << b << ", " << c << ")" << " = " << ans << endl;
    }
    return 0;
}