代码搜索:Num

找到约 10,000 项符合「Num」的源代码

代码结果 10,000
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ucf ps2_e.ucf

#PACE: Start of Constraints generated by PACE #PACE: Start of PACE I/O Pin Assignments NET "b" LOC = "p203" ; NET "CLKIN" LOC = "p182" ; NET "digit" LOC = "p35" ; NET "digit" LOC =
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ucf ps2.ucf

#NET "CLKIN" LOC = "p182" ; #PACE: Start of Constraints generated by PACE #PACE: Start of PACE I/O Pin Assignments NET "DOUT" LOC = "p134" ; NET "DOUT" LOC = "p132" ; NET "DOUT"
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txt 最大报销额-dfs.txt

#include #include using namespace std; double max1,val[26],sum,has[35],max2; int n,num; void dfs(double sum,int p) { if(sum>max1)return; if(pmax2
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cpp 6-4.cpp

#include # define N 10 float average(float *p,int i) { float sum=0.0,aver; for(i=0;i
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c 1004.c

/* This Code is Submitted by icycandy for Problem 1004 on 2008-11-10 at 09:05:41 */ #include /*ID:hrbwy1370451*/ #include /* 函数功能:求回文质数 函数入口参数:整形数A和B 函数返回值:整形数num *
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c 3.c

#include #include #define M 4 #define N 17 #define Myprintf printf("|---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---|\n") /*表格控制*/ typedef struct pa
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m walsh.m

function walsh(m)%m阶阿达玛矩阵求解 % m=m/2;%直积法 % k=[+1 +1;+1 -1]; % c=(log(m)/log(2)); % for i=1:c % k=[k k;k -k]; % end % z=k;%阿达玛矩阵表示 %WALSH码系数相乘法,加入直积法后变为m=m*2 long=length(bz(m-1)); h=zer
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cpp game.cpp

#include #include #include struct Tnum { int a,b; }; Tnum num[210]; int sum[210][210]; int mark[210][210]; int n,m; void init() { int i; scanf("%d%d",&n
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m p3203fai.m

%基本ESPRIT算法2(加入随机相位以后) clear all; N = 128; Pi = 3.1415926; n = [1:1:128]; m= 35; Times=100; %独立仿真次数 f=zeros(Times,2); for k=1:Times w = randn(1,N); fai1=rand*2*Pi-Pi; fai2=rand*
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m p319.m

clear all; N = 128; m =80; Pi = 3.1415926; n = [0:1:N-1]; fsin=0; Times=100; %独立仿真次数 for k=1:Times x = zeros(m,N); for i=1:m rand('state',sum(100*clock)); w = randn