Combinations 组合 思维问题

Combinations

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7499		Accepted: 3509

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

poj上的思维问题。

就是求组合Cnm

队友们竟然提供了三种方法。

1采用 __int64 利用乘一个分子和下面所有的分母进行约分，约分过的就不用继续约分了。这样能有效的减少存储的位数。

#include<stdio.h>
int a[120];
int main()
{
__int64 n,m,i,j,sum=1,num;
while(scanf("%I64d %I64d",&n,&m))
{
if(n==0&&m==0)break;
for(i=1;i<=110;i++)
a[i]=0;
sum=1;
for(i=n-m+1;i<=n;i++)
{
num=i;
for(j=2;j<=m;j++)
{
if(a[j]==1)continue;
if(num%j==0){num/=j;a[j]=1;}

}
sum*=num;
}
for(j=2;j<=m;j++)
{
if(a[j]==0)sum/=j;
}
printf("%I64d things taken %I64d at a time is %I64d exactly.\n",n,m,sum);
}
}

2 利用double来存储长位数，这样知道了分子和分母的总共乘积，就可以求出结果了。

#include<stdio.h>
int main()
{
int n,m,i;
double sum1=1,sum2=1;
while(scanf("%d %d",&n,&m)!=EOF)
{
if(n==0&&m==0)break;
sum1=1;
sum2=1;
for(i=n-m+1;i<=n;i++)
sum1*=double(i);
for(i=1;i<=m;i++)
sum2*=double(i);
printf("%d things taken %d at a time is %.0lf exactly.\n",n,m,sum1/sum2);
}
}

3  杨辉三角 Cnm=Cn-1m+Cn-1m-1

#include<stdio.h>
__int64 a[110][110];
int main()
{
int i,j;
for(i=0;i<=101;i++)
a[i][0]=1;
for(i=1;i<=101;i++)
a[i][i]=1;
for(i=2;i<=101;i++)
for(j=1;j<i;j++)
{
a[i][j]=a[i-1][j]+a[i-1][j-1];
}
while(scanf("%d %d",&i,&j)!=EOF)
{
if(i==0&&j==0)break;
printf("%d things taken %d at a time is %I64d exactly.\n",i,j,a[i][j]);
}

}