<h1 id="heading">题目</h1>
<blockquote>
<h2 id="description">Description</h2>
<blockquote>
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 &lt;= N &lt;= 100; 5 &lt;= M &lt;= 100; M &lt;= 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
</blockquote>
 
<h2 id="input">Input</h2>
<blockquote>
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.
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
The output from this program should be in the form: 
N things taken M at a time is C exactly.
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
100 6 
20 5 
18 6 
0 0
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
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.
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
题目题意比较简单 计算N!/(N-M)!M! 
关键在于数值的计算上 
尽管最后结果我们或许可以保存下，但是其中间要乘到很大的数再除下去，因此要尽可能让中间数小
由于N&gt;M，我们可以剩下很大一部分乘法，只需计算N*(N-1)*······*(M+2)*(M+1)
因此，比较下M和N-M，选择其中较大的与N！约分
然后在计算另一部分，分母和分子同时乘数，每乘一次进行一次约分(gcd)
这样就能在不溢出的情况下计算出我们想要的答案
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;cstring&gt;
#include &lt;cmath&gt;
#include &lt;string&gt;
#include &lt;iostream&gt;
#include &lt;vector&gt;
#include &lt;list&gt;
#include &lt;queue&gt;
#include &lt;stack&gt;
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o&lt;n;o++)

unsigned long long gcd(unsigned long long a, unsigned long long b) {
 return b == 0 ? a : gcd(b, a%b);
}

bool Do() {
 int n, m;
 if (scanf(&#34;%d%d&#34;, &amp;n, &amp;m), n == 0 &amp;&amp; m == 0)
 return false;

 unsigned long long ans = 1;
 int a = max(m, n - m);
 int b = min(m, n - m);
 unsigned long long t = 1;
 for (int i = n, j = 2; i &gt; a; i--, j++) {
 ans *= i;
 if (j &lt;= b || t &gt; 1) {
 if (j &lt;= b)
 t *= j;
 if (t &gt; 1) {
 unsigned long long q = gcd(ans, t);
 ans /= q;
 t /= q;
 }
 }

 }


 printf(&#34;%d things taken %d at a time is %llu exactly.\n&#34;, n, m, ans);

 return true;
}


int main() {
 while (Do());
 return 0;
}
</pre>

POJ 1306.Combinations