uva 11255 Necklace

【题目】

Once upon a time, three girls|Winnie, Grace and Bonnie|owned a large number of pearls. However,
each of them only had a single color of pearls. Winnie had white pearls, Grace had grey pearls and
Bonnie had black pearls. One day, after a long discussion, they decided to make necklaces using the
pearls. They are interested in knowing how many patterns can be formed using a certain number of
pearls of each color, and have asked you to solve this problem for them.
Note that rotating or
ipping over a necklace cannot produce a different kind of necklace. i.e. The
following gure shows three equivalent necklaces.
The following gure shows all possible necklaces formed by 3 white pearls, 2 grey pearls and 1 black
pearl.

Input
The input begins with an integer N ( 2500) which indicates the number of test cases followed. Each
of the following test cases consists of three non-negative integers a, b, c, where 3  a + b + c  40.
Output
For each test case, print out the number of different necklaces that formed by a white pearls, b grey
pearls and c black pearls in a single line.
Sample Input
2
3 2 1
2 2 2
Sample Output
6
11

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【分析】

运用组合数学的知识，根据a+b+c的奇偶性大力讨论。

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【代码】

//UVA 11255 Necklace#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define N 40#define ll long long#define M(a) memset(a,0,sizeof a)#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;const int mxn=45;int n,a,b,c;int C[mxn][mxn];inline int gcd(int x,int y) {return x%y==0?y:gcd(y,x%y);} int main(){    int i,j,T;    fo(i,0,N) C[i][0]=1;    fo(i,1,N)      fo(j,1,i)        C[i][j]=C[i-1][j-1]+C[i-1][j];    scanf("%d",&T);    while(T--)    {        ll ans=0;        scanf("%d%d%d",&a,&b,&c);        n=a+b+c;        fo(i,1,n)        {            int x=n/gcd(n,i);  //循环节大小            if(a%x==0 && b%x==0 && c%x==0)              ans+=C[n/x][a/x]*C[n/x-a/x][b/x];         }        if((n&1) && (a&1)+(b&1)+(c&1)==1)          ans+=C[n/2][a/2]*C[n/2-a/2][b/2];        if(!(n&1))        {             if((a&1)+(b&1)+(c&1)==0)              ans+=C[n/2][a/2]*C[n/2-a/2][b/2]*n;            else if((a&1)+(b&1)+(c&1)==2)              ans+=C[n/2-1][a/2]*C[n/2-1-a/2][b/2]*n;        }        printf("%lld\n",ans/n/2);    }    return 0;}