Arif in Dhaka (First Love Part 2) [Uva 10294]

题目地址请点击——

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Arif in Dhaka (First Love Part 2)

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Description [English]

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Our hero Arif is now in Dhaka (Look at problem 10244 – First Love if you want to know more about Arif, but that information is not necessary for this problem. In short, Arif is a brilliant programmer working at IBM) and he is looking for his first love. Days pass by but his destiny theory is not working anymore, which means that he is yet to meet his first love. He then decides to roam around Dhaka on a rickshaw (A slow vehicle pulled by human power), running DFS (by physical movement) and BFS (with his eyes) on every corner of the street and market places to increase his probability of reaching his goal. While roaming around Dhaka he discovers an interesting necklace shop. There he finds some interesting necklace/bracelet construction sets. He decides to buy some of them, but his programmer mind starts looking for other problems. He wants to find out how many different necklace/bracelet can be made with a certain construction set. You are requested to help him again. The following things are true for a necklace/bracelet construction set.
a) All necklace/bracelet construction sets has a frame, which has N slots to place N beads.
b) All the slots must be filled to make a necklace/bracelet.
c) There are t types of beads in a set. N beads of each type are there in the box. So the total number of beads is t · N (t multiplied by N), of which exactly N can be used at a time.

The figure above shows necklaces for some different values of N (Here, t is always 2). Now let’s turn out attentions to bracelets. A bracelet is a necklace that can be turned over (A junior programmer in Bangladesh says that wrist watch is a necklace (Boys!!! Don’t mind :-))). So for a bracelet the following two arrangements are equivalent. Similarly, all other opposite orientation or mirror images are equivalent.

So, given the description of a necklace/bracelet construction set you will have to determine how
many different necklace and bracelet can be formed with made with that set

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Description [Chinese]

项链和手镯都是由若干珠子串成的环形首饰，区别在于手镯可以翻转，但项链不可以。
换句话说，下面两个图，如果是手镯则看做相同，项链则看做不同。
当然，不管是项链还是手镯，旋转之后一样的看做相同。

输入整数 n 和 t，输出用 t 种颜色的 n 颗珠子（每种颜色的珠子个数无限制，但珠子总数必须是 n）能制作成的项链和手镯的个数。比如 n=5，t=3 时，项链有 51 个，手镯有 39 个。

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Input

The input file contains several lines of input. Each line contains two positive integers N (0 < N < 51) and t (0 < t < 11) as described in the problem statement. Also note that within this input range inputs will be such that no final result will exceed 11 digits. Input is terminated by end of file.

输入包含多组数据。每组数据仅一行，包含两个整数 n 和 t（1<=n<=50，1<=t<=10）。
输入结束标志为文件结束符（EOF）。

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Output

For each line of input produce one line of output which contains two round numbers NN and NB separated by a single space, where NN is the number of total possible necklaces and NB is the number of total possible bracelets for the corresponding input set.

对于每组数据，输出项链的个数和手镯的个数。

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Sample Input

5 2
5 3
5 4
5 5

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Sample Output

8 8
51 39
208 136
629 377

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Solution

根据 Pólya 定理，

f项链=∑i=1ntgcd(i,n)÷n

f手镯={(ntn+12+f项链⋅n)÷2n(n2(tn/2+1+tn/2)+f项链⋅n)÷2nn≡1(mod 2),n≡0(mod 2).

我惊讶地发现，竟然没有打高精度就过了……

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Code

#include <iostream>#include <cstdio>#define LL long longusing namespace std;LL n,t,power[101];LL gcd(LL x,LL y){    return y==0?x:(gcd(y,x%y));}int main(){    while(scanf("%lld%lld",&n,&t)!=EOF){        LL ta=0,tb=0;        power[0]=1;        for(LL i=1;i<=n;i++)power[i]=power[i-1]*t;        for(LL i=1;i<=n;i++)ta+=power[gcd(i,n)];        if(n&1)tb=n*power[(n+1)/2];        else tb=(n*(power[n/2+1]+power[n/2]))/2;        printf("%lld %lld\n",ta/n,(ta+tb)/2/n);    }    return 0;}