Choose and divide UVA - 10375

最新推荐文章于 2020-02-18 01:10:59 发布

原创最新推荐文章于 2020-02-18 01:10:59 发布 · 241 阅读

0 ·

CC 4.0 BY-SA版权

数论同时被 2 个专栏收录

20 篇文章

订阅专栏

UVA-LA算法题目

16 篇文章

订阅专栏

本文详细探讨了UVA在线判题系统中的第10375题，重点讲解如何有效地进行'Choose and divide'策略分析，通过实例解析算法思路，帮助读者理解并解决此类问题。

Choose and divide

UVA - 10375

The binomial coefficient C(m, n) is defined as C(m, n) = m! / (m − n)! n!
Given four natural numbers p, q, r, and s, compute the the result of dividing C(p, q) by C(r, s).
Input
Input consists of a sequence of lines. Each line contains four non-negative integer numbers giving values
for p, q, r, and s, respectively, separated by a single space. All the numbers will be smaller than 10,000 with p ≥ q and r ≥ s.
Output
For each line of input, print a single line containing a real number with 5 digits of precision in the fraction,
giving the number as described above. You may assume the result is not greater than 100,000,000.
Sample Input
10 5 14 9
93 45 84 59
145 95 143 92
995 487 996 488
2000 1000 1999 999
9998 4999 9996 4998
Sample Output
0.12587
505606.46055
1.28223
0.48996
2.00000
3.99960

题意：已知C(m,n) = m!/(n!(m-n)!)，输入整数p,q,r,s,计算C(p,q)/C(r,s).输出保证不超过10^8,保留五位小数。

题解：一道需要用到唯一分解定理的题目，数据范围是10000，先将10000的质数求出，然后统计最后的唯一分解式中每一个素数的指数即可。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

const int maxn = 10005;

int prime[maxn];
int isprime[maxn];
int k;
int n,m,p,q;
int e[maxn];
void intc()
{
    k = 0;
    memset(e,0,sizeof(e));
    memset(prime,0,sizeof(prime));
    for(int i=2;i<=maxn;i++)
    {
        if(prime[i]==0)isprime[k++]=i;
        for(int j=i*2;j<=maxn;j+=i)prime[j]=1;
    }
}
void get_ans(int n,int d)
{
    for(int i=0;i<k;i++)
    {
        while(n%isprime[i]==0)
        {
            n/=isprime[i];
            e[i]+=d;
        }
    }
}
void add_integer(int n,int d)
{
    for(int i=1;i<=n;i++)get_ans(i,d);
}
int main()
{
    while(scanf("%d %d %d %d",&n,&m,&p,&q)!=EOF)
    {
        intc();
        add_integer(n,1);
        add_integer(m,-1);
        add_integer(n-m,-1);
        add_integer(p,-1);
        add_integer(q,1);
        add_integer(p-q,1);
        double ans = 1;
        for(int i=0;i<k;i++)
        {
            ans = ans *pow(isprime[i],e[i]);
        }
        printf("%.5f\n",ans);
    }
    return 0;
}