POJ 题目1306 Combinations（排列组合）

Combinations

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8701		Accepted: 4047

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

UVA Volume III 369

排列组合公式

c(n,a)=c(n-1,a-1)+c(n-1)(a);

ac代码

#include
#include
__int64 c[110][110];
void fun()
{
	int i,j;
	for(i=0;i<=100;i++)
	{
		c[i][0]=1;
	}
	for(i=1;i<=100;i++)
		for(j=1;j<=100;j++)
			c[i][j]=c[i-1][j-1]+c[i-1][j];
}
int main()
{
	int a,b;
	fun();
	while(scanf("%d%d",&a,&b)!=EOF,a||b)
	{
		printf("%d things taken %d at a time is %I64d exactly.\n",a,b,c[a][b]);
	}
}