C. The world is a theatre

Time limit per test

2 seconds

Memory limit per test

256 megabytes

Input

Standard Input

Output

Standard output

There are n boys and m girls
attending Theater Club. to set a play "The Big Bang Theory", they need to choose a group containing exactly T actors containing no less
than 4 boys and no less than one girl. how many ways are there to choose a group? Of course, the variants that only differ in the composition of the troupe are considered different.

Perform all calculations in the 64-bit type:Long longFor tables/Tables ++,Int64For
Delphi andLongFor Java.

Input

The only line of the input data contains three IntegersN,M,T(4 bytes ≤ bytesNLimit ≤ Limit 30, limit 1 limit ≤ limitMLimit ≤ Limit 30, limit 5 limit ≤ limitTLimit ≤ limitNRegion + RegionM).

Output

Find the required number of ways.

Please do not use the % LLD specificator to read or write 64-bit integers in memory ++. It is preferred to use Cin, cout streams or the % i64d specificator.

Sample test (s) Input

 
5 2 5
 

Output

10
 

Input

 
4 3 5
 

Output

3
 

Problem solving Description: This is a typical DP problem. According to the solution of the DP problem, it is nothing more than making a table first and then determining the specified location. In this question, we use C [I] [J] to represent the number of candidates for selecting J as an actor from I, so we can obtain the following transfer equation:

C [I] [J] = C [I-1] [J-1] + C [I-1] [J];

That is, either select the I-1 from the J-1 individual, plus the first I; or select the J from the I-1. The final result is to count the sum of C [N] [I] * C [m] [t-I] from I = 4 until the end of the T-1 [I] boys and girls]

# Include <iostream> # include <cstdio> # include <cstdlib> # include <cmath> # include <cstring> # include <string> # include <set> # include <algorithm> using namespace STD; long long C [64] [64]; long res; int main () {int n, m, T; int I, J; scanf ("% d", & N, & M, & T); for (I = 0; I <= 30; I ++) {c [I] [0] = 1; for (j = 1; j <= I; j ++) {c [I] [J] = C [I-1] [J-1] + C [I-1] [J] ;}res = 0; for (I = 4; I <t; I ++) {res + = C [N] [I] * C [m] [t-I];} printf ("% LLD \ n ", res); Return 0 ;}