代码搜索:Matrix

找到约 10,000 项符合「Matrix」的源代码

代码结果 10,000
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c dlatb4.c

/* -- translated by f2c (version 19940927). You must link the resulting object file with the libraries: -lf2c -lm (in that order) */ #include #include "f2c.h" /* Table of constant
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c zlatm2.c

/* -- translated by f2c (version 19940927). You must link the resulting object file with the libraries: -lf2c -lm (in that order) */ #include "f2c.h" /* Double Complex */ VOID zlatm2_(doublec
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c zlatb4.c

/* -- translated by f2c (version 19940927). You must link the resulting object file with the libraries: -lf2c -lm (in that order) */ #include #include "f2c.h" /* Table of constant
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m bestserverularea.m

%BESTSERVERULAREA [NEEDEDMSTXPOWER BESTSERVER DOMINANCE] = BESTSERVERVULAREA(PERF, % LINKLOSS, MSMAXTXPOW, SPEED, BITRATE, ACTIVESETT) % calculates dominance,
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m sizem.m

function s = sizem(m, dims) % DESCRIPTION res = sizem(m, dims) % For a matrix m the size is given for each element of dims. % Dims is the dimensions for which the size is determined. % If dims is
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cpp find the mincost route(最短路).cpp

//问题是求最小花费的环路(注意是环) //枚举每条边,先去掉 //再value = Dijkstra(s,t)求出最短路 //判断 + value是否最小即可 #include #include int matrix[110][110],path[110],n,m; //matrix[][], 3
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c 逆矩阵.c

#define N 5 /*[注]:修改6为你所要的矩阵阶数*/ #include "stdio.h" #include "conio.h" /*js()函数用于计算行列式,通过递归算法实现*/ int js(s,n) int s[][N],n; {int z,j,k,r,total=0; int b[N][N];/*b[N][N]用于存放,在矩阵s[
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html demosimilaritygraphs.html

GraphDemo
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h dgl.h

//----------------------------------------------------------------------------- // Torque Game Engine // Copyright (C) GarageGames.com, Inc. //------------------------------------------------------
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m fratfd.m

function [r, m] = fratfd(f) % FRATFD Folded Finite Radon Transform % % [r, m] = fratfd(f) % % Input: % f: a N by N matrix where (2N-1) is a prime number. % % Output: %