Combinations
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 10205 | Accepted: 4650 |
Description
Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following:
GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N
Compute the EXACT value of: C = N! / (N-M)!M!
You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long. For the record, the exact value of 100! is:
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000
Input
The input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M pair with both values equal to zero. Your program should terminate when this line is read.
Output
The output from this program should be in the form:
N things taken M at a time is C exactly.
Sample Input
100 6
20 5
18 6
0 0
Sample Output
100 things taken 6 at a time is 1192052400 exactly.
20 things taken 5 at a time is 15504 exactly.
18 things taken 6 at a time is 18564 exactly.
Source
m'ban
模板题,求组合数汇总(点这里)
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
using namespace std;
typedef long long ll;
ll a[100000000];
ll C(ll n,ll 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;
}
return a[m];
}
int main()
{
ll n,m;
while (~scanf("%lld%lld",&n,&m) && (n || m))
{
printf("%lld things taken %lld at a time is %lld exactly.\n",n,m,C(n,m));
}
}