20141002 【数论——组合数】 UVa 10375 Choose and divide

最新推荐文章于 2020-02-21 21:52:14 发布

ZONE画派

最新推荐文章于 2020-02-21 21:52:14 发布

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分类专栏：模板数论 UVa

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本文介绍了一个计算两个组合数比值的问题，通过最小化乘除操作来提高计算效率，并给出了解决方案的代码实现。

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E. Choose and divide

Time Limit: 3000ms

Memory Limit: 131072KB

64-bit integer IO format: %lld Java class name: Main

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Problem D: Choose and divide

The binomial coefficient C(m,n) is defined as

         m!
C(m,n) = --------
         n!(m-n)!

Given four natural numbers p, q, r, and s, compute the the result of dividing C(p,q) by C(r,s).

The 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.

The 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

Output for Sample Input

0.12587
505606.46055
1.28223
0.48996
2.00000
3.99960

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不得不说，这题极度脑残了。

明明长度是 min(p-q, q)，我偏偏写成 min(p-q+1, q)；

结果死的不能再死。。。

中途改用一大堆方法，都没搞定；

以为是精度问题。。。（题目就有说明了...）

还自己给自己出了一组超坑爹的数据。o(╯□╰)o

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <time.h>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <iterator>
using namespace std;
typedef long long LL;
const int MOD = 1e9 + 10;
const int MAXN = 1e6 + 10;

int n, m;
int p, q, r, s;

int main(){     
    while( EOF != scanf("%d%d%d%d", &p, &q, &r, &s) ){
        int lenq = min(p-q, q);
        int lenr = min(r-s, s);
        int len = max(lenq, lenr);
        double ans = 1.0;
        for(int i=0; i<len; i++){
            if( i<lenq )    ans = ans*(p-i)/(i+1);
            if( i<lenr )    ans = ans/(r-i)*(i+1);
        }
        printf("%.5lf\n", ans);
    }
    return 0;
}