Consider how our digital content consumption works today. Some people consume free music and TV shows daily via pirated files, illegal streams. Some consume digital content via legal, publicly-funded radio and TV station streams. Others pay for the privilege to access highly curated and secure services like Spotify, Pandora, Netflix, Hulu, and others. AR clouds will be no different.So how will we choose? I think the answer is security and convenience.
voiddecord(){ cout<<"\n译码: "<<endl; double res = 0.0;double w = 0.5; for(int i=0;i<l;i++,w*=0.5){ res += w*(ans[i]-'0'); } cout<<"译码选取的数字为:"<<res<<endl; int a=0,b=0; // 用a,b,c来记录下一个概率分布数组 double p = 1; //还是需要记录一个p(概率长度),因为这里的table是通用映射表，需要加约束 t.resize(0); double be = 0; for(int i=0;i<sl;i++){ // 2019年4月27日20:52:03 发现还是不能仅跟长度对比啊，没有be，end不行 if(i==0){// 记住左闭右开 if(res<1/3.0){t+='a';a = 1;be = 0.0;} elseif(res<2/3.0){t+='b'; a = 2; be = 1/3.0;} else {t+='c';a = 3; be = 2/3.0;} p = 1/3.0; } elseif(i==1){ if(res < be + p*table[a][1][0]){//be 不变 t += 'a'; b = 1; p *= table[a][1][0]; } elseif(res < be + p*(table[a][2][0]+table[a][1][0])){ t += 'b'; b = 2; be += p*(table[a][1][0]); // be += p*table[a][2][0]; // 2019年4月27日21:49:18今天的bug也很抠脚 p *= table[a][2][0]; } else{ t += 'c'; b = 3; be += p*(table[a][1][0]+table[a][2][0]); p *= table[a][3][0]; } } else{// eg: c/ab [1][2][3] c/a [1][3][0] a [1][0][0]自己开始的假设碰巧有一致性，下次还是要打草稿 // 要先p*=table[a][b][1]再更新a,b if(res < be + p*table[a][b][1]){ t += 'a'; p*=table[a][b][1];// 更新a,b的值 a = b; b = 1; } elseif(res < be + p*(table[a][b][2]+table[a][b][1])){ t += 'b'; be += p*table[a][b][1]; p*=table[a][b][2]; a = b; b = 2; } else{ t += 'c'; be += p*(table[a][b][2]+table[a][b][1]); p*=table[a][b][3]; a = b; b = 3; } } } cout<<"译码结果为:"<<t<<endl; }
#include<bits/stdc++.h> usingnamespace std; #define pb push_back #define x first #define y second string s, ans, t; int num[26]; int Myhash[26]; char Myhash2[26]; int XNlen; typedef pair<double, pair<double, double>> Node; vector<Node> p;
intmain(int argc, charconst *argv[]) { ios::sync_with_stdio(false); cout<<"请输入需要编码的字符序列(编码符号为小写哦):"<<endl; while (cin >> s&&s[0]!='#') { XNlen = 0; p.clear(); memset(num, 0, sizeof(num)); memset(Myhash, 0, sizeof(Myhash)); memset(Myhash2, 0, sizeof(Myhash2)); int n = (int)s.length(); for (int i = 0; i < n; i++) { num[s[i] - 'a']++; }
// Init // 先建立通用的概率，然后再进行讨论 // pair<possibility,<be,end>> + hash[char] = order double be = 0, end = 0; int k = 0; double temd; for (int i = 0; i < 26; i++) { if (num[i]) { // 不为0说明有值的 be = end; temd = (double)num[i] / n; end = be + temd; XNlen += ceil(log(1 / temd) / log(2)) * num[i]; p.pb({temd, {be, end}}); Myhash[i] = k; Myhash2[k] = 'a' + i; //在vscode debug结果令人震惊的时候还是得手动 // cout<<"be: "<<be<<" end: "<<end<<endl; k++; } } // p的构建好像出了问题，2019年4月28日12:34:18 来测试一下 最后发现没有出错... // for (int i = 0; i < (int)p.size();i++){ // cout << "p[i].x:" << p[i].x << " p[i].y.x:" << p[i].y.x << " p[i].y.y:" << p[i].y.y << endl; // }
// encode double pos = 1; be = 0, end = 0; for (int i = 0; i < n; i++){ int tem = Myhash[s[i] - 'a']; be += pos * p[tem].y.x; // end += pos * p[tem].y.y; 昨天的bug 今天又错了，就是不能这样操作因为可能第一次就有end = 1，可以不维护end,可以的 pos *= p[tem].x; // cout << "tem:" << tem << " p[tem].y.x" << p[tem].y.x << " p[tem].y.y:" <<p[tem].y.y << endl; // cout << "be:" << be << " end:" << end << endl; // cout << "pos监控: " << pos << endl; } end = be + pos; // cout << "be:" << be << " end:" << end << endl; // cout << "pos监控: " << pos << endl; int l = ceil(log(1 / pos) / log(2)); double tp = 0.01 * be + 0.99 * end; int tp2; ans.resize(0); for (int i = 0; i < l; i++) { tp *= 2; tp2 = (int)tp; tp -= tp2; ans += char(tp2 + '0'); } cout << "编码长度为: " << l << " 香农编码长度为:" << XNlen << endl; if (l < XNlen) { cout << "WOW,Algorithm Code 果然名不虚传！" << endl; } cout << "编码结果为: " << ans << endl; int sl = int(s.length());
// 译码 cout << "\n译码: " << endl; double res = 0.0; double w = 0.5; for (int i = 0; i < l; i++, w *= 0.5) { res += w * (ans[i] - '0'); } cout << "译码选取的数字为:" << res << endl; pos = 1; t.resize(0); be = 0; for (int i = 0; i < sl; i++) { for (int j = 0; j < k; j++) { // cout << "p[j].y.x:" << pos*p[j].y.x << "p[j].y.y" << pos*p[j].y.y << endl; // if (res >= pos * p[j].y.x && res < pos * p[j].y.y) // mdzz,这个p值会越来越小，所以根本不能用 这样的比较，必须还是要和be和end比较 1. 没有借鉴昨天的后果 2. 没有思考清楚参考系的结果 if (res >= be + pos * p[j].y.x && res < be + pos * p[j].y.y) { t += Myhash2[j]; // 这里应该是j对应的反hash be += pos * p[j].y.x; pos *= p[j].x; break;// 不跳出可能再二次迭代出错 } } } cout << "译码结果为: \n" << t << endl; string tstr; tstr += ((s<t)||(s>t)) == 0 ? "恭喜你,译码和源码完全一样哦":"Sorry,可能长度超出了默认精度范围了哦,有空试试高精度吧"; cout<<tstr<<" 此时源码长度为: "<<sl<<endl; cout<<"\n是否验证下一个字符序列,是则输入,否则输入#:"<<endl; } return0; }
#include<bits/stdc++.h> usingnamespace std; #define pb push_back #define x first #define y second string source, ans, t; int num[26]; int Myhash[26]; char Myhash2[26]; int XNlen; typedef pair<double, pair<double, double>> Node; vector<Node> p; int l;
voidsolve(){ int len = int(source.length()); int last = 0,binlast=0;// 一开始没有想到binlast! while(len){ int slen = len >= 8? 8 : len; string s = source.substr(last,slen);
// last += slen; // len -= slen; // 这两个要在最最最后才搞，否则出错 // string s = source.substr(last,slen);// mdzz 写在这 cout<<"s:"<<s<<endl;
p.clear(); memset(num, 0, sizeof(num)); memset(Myhash, 0, sizeof(Myhash)); memset(Myhash2, 0, sizeof(Myhash2)); int n = (int)s.length(); for (int i = 0; i < n; i++) { num[s[i] - 'a']++; }
// Init // 先建立通用的概率，然后再进行讨论 // pair<possibility,<be,end>> + hash[char] = order double be = 0, end = 0; int k = 0; double temd; for (int i = 0; i < 26; i++) { if (num[i]) { // 不为0说明有值的 be = end; temd = (double)num[i] / n; end = be + temd; XNlen += ceil(log(1 / temd) / log(2)) * num[i]; p.pb({temd, {be, end}}); Myhash[i] = k; Myhash2[k] = 'a' + i; k++; } } // p的构建好像出了问题，2019年4月28日12:34:18 来测试一下 最后发现没有出错... // for (int i = 0; i < (int)p.size();i++){ // cout << "p[i].x:" << p[i].x << " p[i].y.x:" << p[i].y.x << " p[i].y.y:" << p[i].y.y << endl; // }
// encode double pos = 1; be = 0, end = 0; for (int i = 0; i < n; i++){ int tem = Myhash[s[i] - 'a']; be += pos * p[tem].y.x; // end += pos * p[tem].y.y; 昨天的bug 今天又错了，就是不能这样操作因为可能第一次就有end = 1，可以不维护end,可以的 pos *= p[tem].x; // cout << "tem:" << tem << " p[tem].y.x" << p[tem].y.x << " p[tem].y.y:" <<p[tem].y.y << endl; // cout << "be:" << be << " end:" << end << endl; // cout << "pos监控: " << pos << endl; } end = be + pos; // cout << "be:" << be << " end:" << end << endl; // cout << "pos监控: " << pos << endl; l = ceil(log(1 / pos) / log(2)); double tp = 0.01 * be + 0.99 * end;
int tp2; // ans.resize(0); for (int i = 0; i < l; i++) { tp *= 2; tp2 = (int)tp; tp -= tp2; ans += char(tp2 + '0'); }
cout << "编码结果为: " << ans << endl; int sl = int(s.length()); cout<<"当前串长:"<<sl<<endl;
// 译码 cout << "\n译码: " << endl; double res = 0.0; double w = 0.5; for (int i = binlast; i < binlast+ l ; i++, w *= 0.5) // 要用0 1 码长搞出来 { res += w * (ans[i] - '0'); }
cout << "译码选取的数字为:" << res << endl; pos = 1; // t.resize(0); be = 0; for (int i = 0; i < sl; i++) { for (int j = 0; j < k; j++) { // cout << "p[j].y.x:" << pos*p[j].y.x << "p[j].y.y" << pos*p[j].y.y << endl; // if (res >= pos * p[j].y.x && res < pos * p[j].y.y) // mdzz,这个p值会越来越小，所以根本不能用 这样的比较，必须还是要和be和end比较 1. 没有借鉴昨天的后果 2. 没有思考清楚参考系的结果 if (res >= be + pos * p[j].y.x && res < be + pos * p[j].y.y) { t += Myhash2[j]; // 这里应该是j对应的反hash be += pos * p[j].y.x; pos *= p[j].x; break;// 不跳出可能再二次迭代出错 } } } cout<<"中间源码为："<<s<<endl; cout<<"中间译码为："<<t.substr(last,slen)<<endl; last += slen; len -= slen; binlast += l; } }