Binomial Showdown
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 18925 | Accepted: 5788 |
Description
In how many ways can you choose k elements out of n elements, not taking order into account?
Write a program to compute this number.
Write a program to compute this number.
Input
The input will contain one or more test cases.
Each test case consists of one line containing two integers n (n>=1) and k (0<=k<=n).
Input is terminated by two zeroes for n and k.
Each test case consists of one line containing two integers n (n>=1) and k (0<=k<=n).
Input is terminated by two zeroes for n and k.
Output
For each test case, print one line containing the required number. This number will always fit into an integer, i.e. it will be less than 2
31.
Warning: Don't underestimate the problem. The result will fit into an integer - but if all intermediate results arising during the computation will also fit into an integer depends on your algorithm. The test cases will go to the limit.
Warning: Don't underestimate the problem. The result will fit into an integer - but if all intermediate results arising during the computation will also fit into an integer depends on your algorithm. The test cases will go to the limit.
Sample Input
4 2 10 5 49 6 0 0
Sample Output
6 252 13983816
Pascal公式
和一些恒等式
用C(n,m) = (n - m + 1) * C(n, m - 1) / m,由于除以m相对没那么容易越界
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
ll a[100000000];
int main()
{
ll n, m;
while (~scanf("%I64d%I64d", &n, &m) && (n || m)) {
if (n - m < m)
m = n - m;
a[0] = 1;
for (int i = 1; i <= m; i++) {
a[i] = (n - i + 1) * a[i - 1] / i;
}
printf("%d\n", a[m]);
}
return 0;
}