代码搜索:全差分

找到约 10,000 项符合「全差分」的源代码

代码结果 10,000
www.eeworm.com/read/398944/7906864

txt 二分查找.txt

#include using namespace std; #define OK 1 #define ERROR -1 #define N 11 /* 数据元素个数 */ #define EQ(a,b) ((a)==(b)) #define LT(a,b) ((a)
www.eeworm.com/read/298817/7933251

c 四分砝码.c

#include #include main() { int weight1,weight2,weight3,weight4,d1,d2,d3,d4,x,flag; printf("The weight is broke up as following 4 pieces:"); for(weight1=1;weight1
www.eeworm.com/read/196894/8052059

m 上证分形分析.m

clear x=xlsread('index'); y=x(1:5:end,2); z1=log(y); z2=0:(length(y)-1); [Coefficients, S_err, XTXI, R_sq, F_val, Coef_stats, Y_hat, residuals, covariance]=mregress(z1, z2', 1); z3=z1-(Coefficie
www.eeworm.com/read/245937/12771453

pdf 二分查找.pdf

www.eeworm.com/read/332124/12777271

c 四分砝码.c

#include #include main() { int weight1,weight2,weight3,weight4,d1,d2,d3,d4,x,flag; printf("The weight is broke up as following 4 pieces:"); for(weight1=1;weight1
www.eeworm.com/read/144160/12811150

txt 分冶法乘法.txt

Re:分治法例题1的解答 参考解答 设X和Y都是n位的二进制整数,现在要计算它们的乘积XY。我们可以用小学所学的方法来设计一个计算乘积XY的算法,但是这样做计算步骤太多,显得效率较低。如果将每2个1位数的乘法或加法看作一步运算,那么这种方法要作O(n2)步运算才能求出乘积XY。下面我们用分治法来设计一个更有效的大整数乘积算法。   图6-3 大整数X和Y的分段  ...
www.eeworm.com/read/245200/12811339

cpp 二分查找.cpp

// 二分查找.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include using namespace std; int main(int argc, char* argv[]) { int total, num, i = 0,
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dsw 二分查找.dsw

Microsoft Developer Studio Workspace File, Format Version 6.00 # WARNING: DO NOT EDIT OR DELETE THIS WORKSPACE FILE! ###############################################################################
www.eeworm.com/read/245200/12811347

dsp 二分查找.dsp

# Microsoft Developer Studio Project File - Name="二分查找" - Package Owner= # Microsoft Developer Studio Generated Build File, Format Version 6.00 # ** DO NOT EDIT ** # TARGTYPE "Win32 (x86) Cons
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plg 二分查找.plg

Build Log --------------------Configuration: 二分查找 - Win32 Debug-------------------- Command Lines Creating temporary file "C:\DOCUME~1\adm\LOCA