Uva10288Coupons

最新推荐文章于 2018-08-24 14:49:57 发布

原创最新推荐文章于 2018-08-24 14:49:57 发布 · 1k 阅读

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本文介绍了一个经典的概率问题——集齐全套优惠券所需的平均箱数，并提供了一种有效的算法实现方案。通过数学公式 n*Σ(1/n) 的应用及适时的分数约分避免了整数溢出的问题，确保了计算的准确性。

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Coupons

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

Coupons in cereal boxes are numbered 1 to n, and a set of one of each is required for a prize (a cereal box, of course). With one coupon per box, how many boxes on average are required to make a complete set of n coupons?

Input

Input consists of a sequence of lines each containing a single positive integer n, 1<=n<=33, giving the size of the set of coupons. Input is terminated by end of file.

Output

For each input line, output the average number of boxes required to collect the complete set of n coupons. If the answer is an integer number, output the number. If the answer is not integer, then output the integer part of the answer followed by a space and then by the proper fraction in the format shown below. The fractional part should be irreducible. There should be no trailing spaces in any line of output.

Sample Input

Sample Output

11 --

58 ------

题目不难

最后答案为 n*Zegma(1,n) 1/i

每次相加都要进行及时的约分避免溢出！

输出很蛋疼！#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;
int n;
LL gcd(LL a,LL b){
    if(b==0)
        return a;
    return gcd(b,a%b);
}
struct fraction{
    LL rator,deno;
    fraction(LL rator=0,LL deno=0):rator(rator),deno(deno){}
    friend fraction operator + (fraction a,fraction b){
        LL t1 = a.rator*b.deno+a.deno*b.rator,t2 = a.deno*b.deno;
        LL g = gcd(t1,t2);
        return fraction(t1/g,t2/g);
    }
};
int getlen(LL a){
    int len = 0;
    while(a){
        a/=10;
        len++;
    }
    return len;
}

int main(){
    while(~scanf("%d",&n)){
        fraction ans=fraction(1,1);
        for(int  i = 2; i <= n; i++){
            ans = fraction(1,i)+ ans;
        }
        LL t = gcd(ans.deno,n);
        ans.deno /= t;
        ans.rator *= n/t;
        LL d = ans.rator/ans.deno,k = ans.rator%ans.deno;
        if(k!=0){
            int lr = getlen(d),lk = getlen(k),ld = getlen(ans.deno);
            for(int i = 0; i <= lr; i++)
                    cout<<" ";
            cout<<k<<endl;
            cout<<d<<" ";
            for(int i = 0; i < ld; i++)
                cout<<"-";

            cout<<endl;
            for(int i = 0; i <= lr; i++)
                cout<<" ";
            cout<<ans.deno<<endl;
        }else{
            cout<<d<<endl;
        }
    }
    return 0;
}