2020.02.05

Author: ultraji

huawei/0042.cpp

@@ -0,0 +1,64 @@
+// 0042.学英语
+
+#include <iostream>
+using namespace std;
+
+string ge[10] = {"zero", "one", "two", "three", "four", "five", "six", "seven","eight", "nine"};
+string ot[10] = {"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
+string shi[10] = {"zero", "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
+
+string printfHundred(int n)
+{
+    string s = "";
+    bool hasHundred = false;
+    if(n/100 != 0) {
+        s += ge[n/100] + " hundred";
+        hasHundred = true;
+    }
+    n %= 100;
+    if(n != 0) {
+        if (hasHundred) {
+            s += " and ";
+        }
+        if (n < 10) {
+            s += ge[n];
+        }else if (n < 20) {
+            s += ot[n-10];
+        }else {
+            s += shi[n/10];
+            if(n%10){
+                s += " " + ge[n%10];
+            }
+        }
+    }
+    return s;
+}
+
+int main()
+{
+    int n;
+    while (cin >> n)
+    {
+        if (n == 0) {
+            cout << "zero" << endl;
+            continue;
+        }
+        if (n/1000000000 != 0) {
+            cout << ge[n/1000000000] << " billion ";
+        } 
+        n %= 1000000000;
+        if(n/1000000 != 0) {
+            cout << printfHundred(n/1000000) << " million ";
+        }
+        n %= 1000000;
+        if(n/1000 != 0) {
+            cout << printfHundred(n/1000) << " thousand ";
+        }
+        n %= 1000;
+        if(n != 0) {
+            cout << printfHundred(n);
+        }
+        cout << endl;
+    }
+    return 0;
+}

huawei/0044.cpp

@@ -0,0 +1,63 @@
+// 0044.Sudoku-Java
+
+// 测试数据含有非唯一解，不在尝试AC
+
+#include <iostream>
+#include <vector>
+#include <queue>
+#include <map>
+using namespace std;
+
+int v[9][9];
+int lin[9][10], row[9][10], check[9][10];
+
+bool test(int x, int y)
+{
+    int cnt = 0, t = 0, np[10];
+    fill(np, np+10, 0);
+    for(int i = 1; i < 10; i++) {
+        np[i] = lin[x][i] + row[y][i] + check[x/3*3+y/3][i];
+        if(np[i] != 0) {
+            cnt ++;
+        } else {
+            t = i;
+        }
+    }   
+    if(cnt == 8) {
+        v[x][y] = t;
+        lin[x][t] = 1;
+        row[y][t] = 1;
+        check[x/3*3+y/3][t] = 1;
+        return true;
+    }
+
+    return false;
+}
+
+int main()
+{
+    queue<pair<int, int> > q;
+    for(int i = 0; i < 9; i++){
+        for (int j = 0; j < 9; j++){
+            cin >> v[i][j];
+            lin[i][v[i][j]] = 1;
+            row[j][v[i][j]] = 1;
+            check[i/3*3+j/3][v[i][j]] = 1;
+            if(v[i][j] == 0) {
+                q.push(make_pair(i,j));
+            }
+        }
+    }
+    while(!q.empty()) {
+        pair<int, int> p = q.front();
+        q.pop();
+        if (!test(p.first, p.second)) q.push(p);
+    }
+    for(int i = 0; i < 9; i++){
+        for (int j = 0; j < 9; j++){
+            cout << v[i][j] << ' ';
+        }
+        cout << endl;
+    }
+    return 0;
+}

huawei/0045.cpp

@@ -0,0 +1,49 @@
+// 0045.名字的漂亮度
+
+#include <iostream>
+#include <vector>
+#include <map>
+#include <algorithm>
+using namespace std;
+
+int score(string s)
+{
+    map<char,int> mp;
+    vector<int> v;
+    int sum = 0;
+    for(int i = 0; i < s.size(); i++)
+    {
+        if (mp.find(s[i]) == mp.end()) {
+            mp[s[i]] = 1;
+        }else {
+            mp[s[i]] ++;
+        }
+    }
+    for(auto it = mp.begin(); it != mp.end(); it ++)
+    {
+        v.push_back(it->second);
+    }
+    sort(v.begin(), v.end());
+    reverse(v.begin(), v.end());
+    for(int i = 0; i < v.size(); i++)
+    {
+        sum += (26-i)*v[i];
+    }
+    return sum;
+}
+
+int main()
+{
+    int n;
+    string s;
+    while(cin >> n)
+    {
+        getchar();
+        for(int i = 0; i < n; i ++)
+        {
+            getline(cin, s);
+            cout << score(s) << endl;
+        }
+    }
+    return 0;
+}

huawei/0046.cpp

@@ -0,0 +1,19 @@
+// 0046.按字节截取字符串
+
+// 根据题意，在判题系统中，汉字以2个字节存储。
+#include <iostream>
+#include <cstring>
+using namespace std;
+
+int main()
+{
+    string s;
+    int n;
+    while(cin >> s >> n)
+    {
+        if(s[n-1] < 0) s = s.substr(0, n-1);
+        else s = s.substr(0, n);
+        cout << s << endl;
+    }
+    return 0;
+}

huawei/0055.cpp

@@ -0,0 +1,22 @@
+// 0055.（练习用）挑7
+
+#include <iostream>
+using namespace std;
+
+int main()
+{
+    int n = 0;
+    int mp[30001];
+    mp[0] = 0;
+    for(int i = 1; i <= 30000; i++)
+    {
+        string s = to_string(i);
+        if(i%7==0 || s.find('7') != s.npos) mp[i] = mp[i-1]+1;
+        else mp[i] = mp[i-1];
+    }
+    while(cin >> n)
+    {
+        cout << mp[n] << endl;
+    }
+    return 0;
+}

huawei/0056.cpp

@@ -0,0 +1,37 @@
+// 0056.iNOC产品部--完全数计算
+
+// 只计算一遍500000的个数，也容易超时的可能原因：
+// 1. 可能我算法的时间复杂度太大
+// 2. 题目有点问题，说明实际测试的数应该小于明显500000.
+// 实际只有以下4个数：6 28 296 8128
+#include <iostream>
+#include <cmath>
+using namespace std;
+
+bool isPerf(int n)
+{
+    int sum = 1;
+    int t = sqrt(n);
+    for(int i = 2; i <= t; i++){
+        if(n%i == 0) sum += i + n/i;
+    }
+    if(sum == n) return true;
+    return false;
+}
+
+int main()
+{
+    int n = 0;
+    int mp[500001];
+    mp[1] = 0;
+    for(int i = 2; i <= 100001; i++)
+    {
+        if(isPerf(i)) mp[i] = mp[i-1]+1;
+        else mp[i] = mp[i-1];
+    }
+    while(cin >> n)
+    {
+        cout << mp[n] << endl;
+    }
+    return 0;
+}

huawei/0059.cpp

@@ -0,0 +1,29 @@
+// 0059.找出字符串中第一个只出现一次的字符
+
+#include <iostream>
+using namespace std;
+
+int main()
+{
+    string s;
+    int np[128];
+    while(getline(cin, s))
+    {
+        fill(np, np+128, 0);
+        for(int i = 0; i < s.size(); i++) np[s[i]] ++;
+        string r = "";
+        for(int i = 0; i < 128; i++) if(np[i] == 1) r+= (char)i;
+        if (r.empty()) {
+            cout << -1 << endl;
+            continue;
+        }
+        for(int i = 0; i < s.size(); i++)
+        {
+            if(r.find(s[i]) != r.npos){
+                cout << s[i] << endl;
+                break;
+            }
+        }
+    }
+    return 0;
+}