留言板

hnu 2019-3-17 18:16

q=1.7263e-004 %以最低频分量里面大于0.00001绝对值最小的系数作为量化步长,并且这个值作为密钥key保存下来
for i=1:1024
ave(i)=sum(ca3(i*60:(i+1)*60))/60;%因为ca3系数有正亦有负，所以ave(i)∈(-1.0000,11111)正负不定
z(i)=fix(ave(i)/q+1/2);%所以下面z(i)∈(-1000,1000)正负亦不定,q∈(-0.001, 0.001)
end
for i=1:1024
if mod(z(i),2)==w1(i)
cxzc=0;
else
if mod(z(i),2)~=w1(i)&&z(i)==fix(ave(i)/q)&&z(i)>=0||mod(z(i),2)~=w1(i)&&z(i)~=fix(ave(i)/q)&&z(i)<0
ca3(i*60:(i+1)*60)=ca3(i*60:(i+1)*60)+q;
else
ca3(i*60:(i+1)*60)=ca3(i*60:(i+1)*60)-q;
end
end
end
PB1825854387 2019-3-7 16:45

WAHAHA。 2019-3-2 20:14

2. 程序如下：
function yanzheng1(net)
%网络加载,注意文件名要加单引号
inv=[0.550000000000001,0.475000000000001;0.468937500000000,0.964890020864981;1,1;0,0;1,1;0.557000000000000,0.426450000000000;0.961586638830898,0.405219206680585;1,1;0.354100000000000,0.563100000000000;0.0689600000000000,0.0875800000000000;0.551800000000000,0.498900000000000];
%进行仿真
outv=net(inv);
outv(1,:)
end

3.报错信息如下：
C:\Users\lei>C:\Users\lei\Desktop\exe\yanzheng1.exe

> In yanzheng1 (line 3)

MCM_2017 2018-12-25 10:27

ykbin 2018-12-2 14:34

hiahiah_n7pD7 2018-11-29 10:11

Chenle2018 2018-11-10 18:54

net = train(net,x123,y123);
yhat = net(x123);
plotconfusion(y123,yhat);

lss1103 2018-10-18 16:58

hold on
plot(x3,y2)