Cattelan number + large number

The "Harry Potter and the Goblet of Fire" will be on show in the next few days. As a crazy fan of Harry Potter, you will go to the cinema and have the first sight, won’t you?

Suppose the cinema only has one ticket-office and the price for per-ticket is 50 dollars. The queue for buying the tickets is consisted of m + n persons (m persons each only has the 50-dollar bill and n persons each only has the 100-dollar bill).

Now the problem for you is to calculate the number of different ways of the queue that the buying process won't be stopped from the first person till the last person.
Note: initially the ticket-office has no money.

The buying process will be stopped on the occasion that the ticket-office has no 50-dollar bill but the first person of the queue only has the 100-dollar bill.

InputThe input file contains several test cases. Each test case is made up of two integer numbers: m and n. It is terminated by m = n = 0. Otherwise, m, n <=100.
OutputFor each test case, first print the test number (counting from 1) in one line, then output the number of different ways in another line.
Sample Input

3 0
3 1
3 3
0 0

Sample Output

Test #1:
6
Test #2:
18
Test #3:
180 

question meaning: The classic Cattelan number problem, the final answer is C(m+n, m) - C(m+n, m+1) , and then further simplification can be obtained (m+n)!(m+ 1-n)/(m+1) , a large number multiplication is OK 
code example:

#define ll long long
const int maxn = 1e6+5;
const int mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

int m, n;
vector<int>a, b, c;
stack<int>s;

void bigsum(){
    c.clear();
    c.assign(a.size()+b.size()-1, 0);
    for(int i = 0; i < b.size(); i++){
        int k = i;
        for(int j = 0; j < a.size(); j++){
            c[k] += b[i]*a[j];
            k++;
        }
    }
    
    for(int i = c.size()-1; i >= 0; i--){
        if (c[i] > 9){
            if (i != 0){c[i-1] += c[i]/10; c[i] %= 10;}
            else {
                int tem = c[i]/10;
                c[i] %= 10;
                c.insert(c.begin(), tem);
            }
        }
    }
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int case = 1;
    
    while(~scanf("%d%d", &m, &n) && n+m){
        printf("Test #%d:\n", kase++);
        if (m < n) printf("0\n");
        else{
            b.clear();
            b.push_back(1);
            for(int i = 2; i <= m+n; i++){
                if (i == m+1) continue;
                a.clear();
                int x = i;
                while(x) s.push(x%10), x /= 10;
                while(!s.empty()){
                    a.push_back(s.top());
                    s.pop();
                }
                bigsum();
                b = c;       
            }
            
            a.clear();
            int x = m+1-n;
            if (x != m+1) {
                while(x) s.push(x%10), x /= 10;
                while(!s.empty()){
                    a.push_back(s.top());
                    s.pop();
                }
                bigsum();
            }
            for(int i = 0; i < c.size(); i++){
                printf("%d", c[i]);
            }
            printf("\n");
        }
    }
    return 0;
}