Combinations
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 11049 | Accepted: 5013 |
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.
题意：
输入n,k，然后算一下组合数就行了，关键：“You may assume that the final value of C will fit in a 32-bit”，所以还是很无聊的一题。
AC code:
#include<cstdio>
using namespace std;
typedef long long ll;
ll c[110][110];
void prepare()
{
for(int i=0;i<=110;i++) c[i][0]=1;
for(int i=1;i<=110;i++)
for(int j=1;j<=110;j++)
c[i][j]=c[i-1][j]+c[i-1][j-1];
}
int main()
{
//freopen("input.txt","r",stdin);
prepare();
ll n,k;
while(~scanf("%lld%lld",&n,&k)&&n)
{
printf("%lld things taken %lld at a time is %lld exactly.\n",n,k,c[n][k]);
}
return 0;
}
原文地址：https://www.cnblogs.com/cautx/p/11404467.html