diff --git a/GPLT/CPP/L2-028 秀恩爱分得快 (25 分).cpp b/GPLT/CPP/L2-028 秀恩爱分得快 (25 分).cpp
new file mode 100644
index 0000000..b08d960
--- /dev/null
+++ b/GPLT/CPP/L2-028 秀恩爱分得快 (25 分).cpp	
@@ -0,0 +1,92 @@
+分析：1.此题考查的是对 0 -0 的特殊处理。当遇到 0 时候把此人转为1000存储, 所以读数据的时候要以字符串形式读取
+2.遍历每张照片，把与男主女主对应的异性亲密度用sum数字累加起来， 并维护maxn[1] maxn[2]， 为男主女主的最亲密值 和ans[1], ans[2]容器，为最亲密异性id。
+3.判断男主女主是否互为最亲密，如果是，输出并return 0， 否则分别输出他们的最亲密好友
+4.注意输出时候，因为把0当1000存储，会导致0号人排在最后，这是不符题意的，输出之前排个序，让1000排在最前面
+
+#include <iostream>
+#include <cmath>
+#include <vector>
+#include <algorithm>
+using namespace std;
+bool cmp (int a, int b) {
+    if(abs(a) == 1000) return true;
+    if(abs(b) == 1000) return false;
+    return abs(a) < abs(b);
+}
+int main(){
+    int n, m, num, k, sex[1010] = {0}, love[3];
+    double sum[1010] = {0}, maxn[3] = {0} ;
+    string s;
+    cin >> n >> m;
+    vector<vector<int>> v(m), ans(3);
+    for(int i = 0; i < m; i++) {
+        cin >> k;
+        for(int j = 0; j < k; j++){
+            cin >> s;
+            if(s == "0")  s = "1000";
+            if(s == "-0") s = "-1000";
+            num = stoi(s);
+            sex[abs(num)] = num;
+            v[i].push_back(num);
+        }
+    }
+    for(int i = 1; i <= 2; i++) {
+        cin >> s;
+        if(s == "0")  s = "1000";
+        if(s == "-0") s = "-1000";
+        love[i] = stoi(s);
+    }
+    for(int k = 1; k <= 2; k++) {
+        for(int i = 0; i < m; i++) {
+            int flag = 0;
+            for(int j = 0; j < v[i].size(); j++){
+                if(v[i][j] == love[k]) {
+                    flag = 1;
+                    break;
+                }
+            }
+            if(flag == 1) {
+                for(int j = 0; j < v[i].size(); j++){
+                    if(love[k] * v[i][j] < 0)  {
+                        sum[(int)abs(v[i][j])] += 1.0 / v[i].size();
+                    }
+                }
+            }
+        }
+    }
+    maxn[1] = maxn[2] = -1;
+    for(int k = 1; k <= 2; k++) {
+        for(int i = 1; i <= 1000; i++) {
+            if(love[k] * sex[i] < 0) {
+                if(sum[i] > maxn[k]) {
+                    maxn[k] = sum[i];
+                    ans[k].clear();
+                    ans[k].push_back(sex[i]);
+                }else if(sum[i] == maxn[k]) {
+                    ans[k].push_back(sex[i]);
+                }
+            }
+        }
+    }
+    if(maxn[1] == sum[(int)abs(love[2])] && maxn[2] == sum[(int)abs(love[1])]) {
+        string s1 = to_string(love[1]), s2 = to_string(love[2]);
+        if(love[1] == 1000)  s1 = "0";
+        if(love[1] == -1000)  s1 = "-0";
+        if(love[2] == 1000)  s2 = "0";
+        if(love[2] == -1000)  s2 = "-0";
+        cout << s1 << " " << s2 << endl;
+        return 0;
+    }
+    for(int k = 1; k <= 2; k++) {
+        sort(ans[k].begin(), ans[k].end(), cmp);
+        for(int i = 0; i < ans[k].size(); i++) {
+            string s1 = to_string(love[k]), s2 = to_string(ans[k][i]);
+            if(love[k] == 1000)  s1 = "0";
+            if(love[k] == -1000)  s1 = "-0";
+            if(ans[k][i] == 1000)  s2 = "0";
+            if(ans[k][i] == -1000)  s2 = "-0";
+            cout << s1 << " " << s2 << endl;
+        }
+    }
+    return 0;
+}