排列组合计数&POJ306

题意：

计算C = N! / (N-M)!M!

方法一：

#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;

typedef long long LL;
LL solve(LL n, LL k)
{
    if(k >n /2)
        k = n-k;
    LL a = 1, b = 1;
    for(int i = 1; i <= k; i++)
    {
        a *= (n + 1 - i);
        b *= i;
        if(a % b == 0)
        {
            a /= b;
            b = 1;
        }
    }
    return a/ b;
}
int main()
{
    //freopen("in.txt", "r", stdin);
    LL n, m;
    while(scanf("%I64d%I64d", &n, &m) && n)
    {
        printf("%I64d things taken %I64d at a time is %I64d exactly.\n", n, m, solve(n, m));
    }
    return 0;
}

方法二：利用二项式系数公式(c[i][j] = c[i - 1][j - 1] + c[i - 1][j])

#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;

typedef long long LL;
const int maxn = 110;
LL c[maxn][maxn];

void init()
{
    for(int  i = 0; i < maxn; i++)
    {
        c[i][0] = 1;
    }
    for(int i = 1; i < maxn; i++)
    {
        for(int j = 1; j <= i; j++)
            c[i][j] = c[i-1][j] + c[i-1][j-1];
    }
}

int main()
{
    //freopen("in.txt", "r", stdin);
    int n, m;
    init();
    while(scanf("%d%d", &n, &m) && n)
    {
        printf("%d things taken %d at a time is %I64d exactly.\n", n, m, c[n][m]);
    }
    return 0;
}