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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. 

InputThe 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.OutputPrint 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

#include
#include
#include

using namespace std;
typedef long long LL;
const int N = 20 + 5;
const int INF = 0x3f3f3f3f;
LL dp[N][N][N];

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  1048576;
    if(dp[a][b][c] != -1) return dp[a][b][c];
    if(a < b && b < c) return dp[a][b][c] = Solve(a, b, c-1) + Solve(a, b-1, c-1) - Solve(a, b-1, c);
    return 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);
}

int main(){
    LL a, b, c;
    memset(dp, -1, sizeof(dp));
    while(scanf("%lld %lld %lld", &a, &b, &c) && (a != -1 || b != -1 || c != -1)){
        LL ans =  Solve(a, b, c);
        printf("w(%lld, %lld, %lld) = %lld\n", a, b, c, ans);
    }
}

 

转载于:https://www.cnblogs.com/Pretty9/p/7406923.html