chess排列组合题 皇星客栈

                             A.Chess

Problem Description

We have a chessboard, as the figure below, it shows the chessboard with dimension 1, 3, 5, 7 (We ignore the even number).Now, your task is to place K bishops on the n – dimension chessboard and satisfied that none of then attack each other ( A bishop can move to any distance in any of the four diagonal directions).

Input

输入包含两个整数n(n < 100 & & n % 2 = = 1)和k(k < = 10000)之间用一个空格分开。

Output

输出的答案mod 100007.

Sample Input

7 3
1 1

Sample Output

1038
1
第一种方法：

#include<stdio.h>

__int64 Q[50];
#define mod 100007

__int64 xiangqi( __int64 n,__int64 k )
{
    __int64 i;
    __int64 j;
    __int64 sum=1;

    for( i = 1; i <= n ; i++ )
        Q[i] = i*i;
    for( i = k; i >=2 ; i-- )
    {    
        j = 1;
        while( i * j <= n )
        {
            if( j*i >= n-k+1 && Q[i*j] % i == 0 )
            {
                Q[i*j] /= i ;
                break;
            }
            j++;
        }
    }
    for( i = n ; i >= n-k+1 ; i-- )
    {
        sum *= Q[i]%mod;
        sum %= mod;
    }
    return sum;
}

    

int main( )
{
    int n;
    int k;
    int num1;
    int num2;
    __int64 sum;
    int i;

    while( scanf("%d %d",&n,&k) == 2 )
    {
        
        sum = 0;
        if( n < 100 && n % 2 == 1 )
        {
            if( k > n )
            { printf("0\n"); continue; }
            if( n == 1 )
            { printf("1\n"); continue; }
            num1 = (n+1)/2;
            num2 = num1 - 1;

            for( i = k; i >= 0 ; i-- )
            {
                if( i > num1 || k - i > num2 )
                    continue;
                sum += xiangqi(num1,i)*xiangqi(num2,k-i)%mod;
                sum %= mod;
            }
        }
        printf("%I64d\n",sum);
    }
    return 0;
}

   第二种方法 ：

#include <iostream>

using namespace std;
#define MOD 100007
int n, k;
__int64 dp[2][101][101];

void DP( int mod, int n ) {

    int i, j;
    for(i = 0; i <= n; i++) {
        for(j = 0; j <= n; j++) {
            dp[ mod ][i][j] = 0;
        }
    }
    dp[ mod ][0][0] = 1;
    for( i = 1; i <= n; i++) {
        dp[mod][i][0] = 1;
        for(j = 1; j <= n; j++) {
            dp[mod][i][j] = dp[mod][i-1][j];
            dp[mod][i][j] += dp[mod][i-1][j-1] * ( n - j + 1 ) % MOD;
            dp[mod][i][j] %= MOD;
        }
    }
}

int main() {
    int i;

    /*freopen ("1005.in", "r", stdin );
    freopen ("1005.out", "w", stdout );*/

    while( scanf("%d %d", &n, &k) != EOF ) {
        DP(0, n/2);
        DP(1, n - n/2);

        if( k > n ) {
            printf("0\n");
            continue;
        }

        __int64 sum = 0;
        for(i = 0; i <= k; i++) {
            if( i > n || k-i > n )
                continue;
            sum = sum + ( dp[0][n/2][i] * dp[1][n-n/2][k-i] ) % MOD;
            sum %= MOD;
        }
        printf("%I64d\n", sum);
    }
    return 0;
}