搜索：N-Queens

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https://www.eeworm.com/dl/617/468502.html 压缩解压

N-Queens made in SST

N-Queens made in SST
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https://www.eeworm.com/dl/641/436763.html 数学计算

n queens problem which can print the chess board pattern

n queens problem which can print the chess board pattern
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https://www.eeworm.com/dl/518/435770.html 数值算法/人工智能

This progamram desrible how put N queens on the board have size N*N Using algorithm back tracking

This progamram desrible how put N queens on the board have size N*N Using algorithm back tracking
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https://www.eeworm.com/dl/654/467179.html 数据结构

This program finds the locations of placing n queens in a n*n chessboard so that none attack any oth

This program finds the locations of placing n queens in a n*n chessboard so that none attack any other
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https://www.eeworm.com/dl/650/474968.html 人工智能/神经网络

C PROGRAM FOR 8 QUEENS PROBLEM USING BACKTRACKING

C PROGRAM FOR 8 QUEENS PROBLEM USING BACKTRACKING
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https://www.eeworm.com/dl/625/156947.html 游戏

Using Genetic Algorithm to solve the 8 Queens problem.

Using Genetic Algorithm to solve the 8 Queens problem.
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https://www.eeworm.com/dl/694/325523.html 并行计算

所谓八皇后问题（Eight Queens Problem）

所谓八皇后问题（Eight Queens Problem），是在8*8格的棋盘上，放置8个皇后。要求每行每列放一个皇后,而且每一条对角线和每一条反对角线上最多只能有一个皇后，这是一个典型并行算法，C语言实现。
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https://www.eeworm.com/dl/654/476993.html 数据结构

Solution to the "8 Queens Problem" (8 Damen Problem) known from Chess

Solution to the "8 Queens Problem" (8 Damen Problem) known from Chess
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https://www.eeworm.com/dl/644/324442.html 汇编语言

系数对称的FIR滤波器设计 * *N=8,h(n)=h(N-1-n) * *y(n)=h0*[x(n)+x(n-7)]+h1*[x(n-1)+x(n-6)] * * +h2*[x(n-

系数对称的FIR滤波器设计 * *N=8,h(n)=h(N-1-n) * *y(n)=h0*[x(n)+x(n-7)]+h1*[x(n-1)+x(n-6)] * * +h2*[x(n-2)+x(n-5)]+h3*[x(n-3)+x(n-4)]
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https://www.eeworm.com/dl/518/385900.html 数值算法/人工智能

//Euler 函数前n项和 /* phi(n) 为n的Euler原函数 if( (n/p) ％ i == 0 ) phi(n)=phi(n/p)*i else phi(n)=phi(n/p

//Euler 函数前n项和 /* phi(n) 为n的Euler原函数 if( (n/p) ％ i == 0 ) phi(n)=phi(n/p)*i else phi(n)=phi(n/p)*(i-1) 对于约数：divnum 如果i|pr[j] 那么 divnum[i*pr[j]]=divsum[i]/(e[i]+1)*(e[i]+2) //最小素因子次数加1 否则 divnum[i*pr[j]]=divnum[i]*divnum[pr[j]] //满足积性函数条件 ...
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