poj 1707 Sum of powers 伯努利数系数

题目大意：

　　对于方程　　   

　　输入 k ( 0 <= k <= 20 )  输出  

 

解题思路：

　　由数学 伯努利数 公式：　　  

　　带入通分化简就可以了，不过要注意，分母为正...( WA了一小时错在了这里 )

　伯努利数是18世纪瑞士数学家雅各布·伯努利引入的一个数。

　　设伯努利数为B(n)，它的定义为：

　　t/(e^t-1）=∑[B(n)*(t^n)/(n!)](n:0->；∞）

　　这里|t|<2。由计算知：

　　B(0)=1,B⑴=-1/2,

　　B⑵=1/6,B⑶=0,

　　B⑷=-1/30,B⑸=0,

　　B⑹=1/42,B⑺=0,

　　B⑻=-1/30,B⑼=0),

　　B⑽=5/66,B⑾=0,

　　B⑿=-691/2730,B⒀=0,

　　B⒁=7/6,B⒂=0,

　　B⒃=-3617/510,B⒄=0,

　　B⒅=43867/798,B⒆=0,

　　B⒇=-174611/330 ……

 

 

解题代码:

View Code

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

typedef long long LL;

const int N = 25;



int B1[N],B2[N];

LL a[N], b[N], C[N][N];



void init(){//预处理伯努利系数

    B1[0] = 1; B2[0] = 1;

    B1[1] = -1;B2[1] = 2;

    B1[2] = 1; B2[2] = 6;

    B1[3] = 0; B2[3] = 0;

    B1[4] = -1; B2[4] = 30;

    B1[5] = 0; B2[5] = 0;

    B1[6] = 1; B2[6] = 42;

    B1[7] = 0; B2[7] = 0;

    B1[8] = -1; B2[8] = 30;

    B1[9] = 0; B2[9] = 0;

    B1[10] = 5,B2[10] = 66;

    B1[11] = 0; B2[11] = 0;

    B1[12] = -691; B2[12] = 2730;

    B1[13] = 0; B2[13] = 0;

    B1[14] = 7; B2[14] = 6;

    B1[15] = 0; B2[15] = 0;

    B1[16] = -3617; B2[16] = 510;

    B1[17] = 0; B2[17] = 0;

    B1[18] = 43867; B2[18] = 798;

    B1[19] = 0; B2[19] = 0;

    B1[20] = -174611; B2[20] = 330;



    // 组合数

    for(int i = 0; i <= 21; i++)

        C[i][0] = C[i][i] = 1;

    for(int i = 1; i <= 21; i++)

    {

        for(int j = 1; j < i; j++)

            C[i][j] = C[i-1][j]+C[i-1][j-1];

    }

}

LL gcd( LL a, LL b ){ return b==0?a:gcd(b,a%b);}

LL lcm( LL a, LL b ){ return (a/gcd(a,b))*b; }



int main(){

    init();

    int k;

    while( scanf("%d", &k) != EOF)

    {

        memset( a, 0, sizeof(a));

        memset( b, 0, sizeof(b));

        

        LL t1[N], t2[N];

        for(int r = 1; r <= k+1; r++)

        {

            memset( t1, 0, sizeof(t1));

            memset( t2, 0, sizeof(t2));    

            for(int i = 0; i <= r; i++)

            {

                t1[i] = B1[k+1-r]*C[k+1][r]*C[r][i];

                t2[i] = B2[k+1-r]; 

            }    

            for(int i = 0; i <= r; i++)

            {

                if( t2[i] ){ 

                    if( b[i] == 0 ){

                        a[i] = t1[i]; b[i] = t2[i];    

                    }    

                    else{//通分

                        LL z = lcm( b[i], t2[i] );

                        LL x = z/b[i], y = z/t2[i];

                        a[i] = a[i]*x + t1[i]*y;

                        b[i] = z;

                    }

                }    

            }

        }

        //获取所有分母最小公倍数 z    

        LL z = 1, p = 1;

        for(int i = 0; i <= k+1; i++)

            if( b[i] )    z = lcm( z, b[i] );

        for(int i = 0; i <= k+1; i++)

            if( b[i] ){ //统一分母z

                a[i] *= (z/b[i]);

                p = a[i]; b[i] = z;    

            }        

        z *= (k+1);    

        //获取分子分母最大公约数 p    

        for(int i = 0; i <= k+1; i++)

            if( b[i] ) p = gcd( p, a[i] );    

        p = gcd( p, z );        

        //简化形式    

        for(int i = 0; i <= k+1; i++)

            if( b[i] ) a[i] /= p;

        

        z /= p;    

        int f = ( z < 0 ) ? -1 : 1;



    //    printf("p = %lld\n", p );

        printf("%lld", f*z);

        for(int i = k+1; i >= 0; i-- )

            printf(" %lld",f*a[i]);

        puts("");

    }

    return 0;

}