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+实验1.1半数集问题
+#include <bits/stdc++.h>
+using namespace std;
+int a[1005]={0}; //存放已计算的数据
+int b[1005]={0};
+int halfset(int n){
+int sum = 1;
+ if(a[n]>0) //如果f(n)已经得出，就不必再重复计算
+        return a[n];
+    for(int i=1;i<=n/2;i++){
+        sum += halfset(i);
+    }
+    a[n]=sum;
+    return sum;
+}
+int main()
+{
+    int n = 0,i = 0;
+	cin>>n;
+	while(n!=0){
+		b[i++] = n;
+		cin>>n;
+	}
+	for(int j=0;j<i;j++){
+		int result=halfset(b[j]);
+		cout << result <<endl;
+	}
+    return 0;
+}
+实验1.2双色hanoi塔问题
+#include <bits/stdc++.h>
+#include<cstdio>
+#include<cstring>
+void mov(int n,char a,char b)
+{
+	printf("%d %c %c\n",n,a,b);
+}
+void dg(int n,char a, char c, char b)
+{
+	if(n==1)  mov(n,a,b);
+	else
+	{
+		dg(n-1,a,b,c);
+	    mov(n,a,b);
+		dg(n-1,c,a,b);
+	}
+}
+int main()
+{
+    char  a='A',b='B',c='C';
+    int n;
+	while(scanf("%d",&n)!=EOF)
+        dg(n,a,c,b);
+	return 0;
+}
+实验1.3 整数因子分解问题
+#include <bits/stdc++.h>
+using namespace std;
+map<int ,int >a;
+int f(int n){
+   if(n == 1)
+       return 1;
+   if(a[n])
+       return a[n];
+   int count=1;
+   for(int i=2;i<=sqrt(n);i++){
+   	       if(n%i == 0){
+           count+=f(i);
+           if(i!=n/i){
+               count+=f(n/i);
+           }
+       }
+   }
+   a[n] = count;
+   return a[n];
+}
+int main()
+{
+   int n;
+   while(cin>>n){
+       cout<<f(n)<<endl;
+   }
+}
+
+
+实验2.1我要去西藏
+#include<bits/stdc++.h>
+#include<algorithm>
+using namespace std;
+int a[200][200];
+int cost[200];
+int city,result;
+int travel(){
+	cost[city-2] = a[city-2][city-1];
+	for(int i = city-3;i>=0;i--){
+		result = a[i][city-1];
+		for(int j = i+1;j<city-1;j++){
+			result = min(a[i][j]+cost[j],result);
+		}
+		cost[i] = result;
+	}
+	return cost[0];
+}
+int main(){
+	cin>>city;
+	for(int i = 0;i<city-1;i++){
+		for(int j = i+1;j<city;j++){
+			cin>>a[i][j];
+		}
+	}
+	result = travel();
+	cout<<result;
+	return 0;
+}
+实验2.2云天明项链
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 1e3 + 10;
+#define Ma_x 99999
+#define Mi_x 0 
+#define min(a,b) a<b?a:b
+#define max(a,b) a>b?a:b 
+int n,w[N],dp[N][N],dq[N][N];
+int sum(int i,int t){
+    int k,s=0,k1;
+    for(k=i;k<i+t;k++){
+        k1=k%n;
+        if(k1==0) k1=n;
+        s=s+w[k1];   
+    }
+    return s;
+}
+int main(){
+    int i,t,k;
+    scanf("%d",&n);
+    for(i=1;i<=n;i++){
+        scanf("%d",&w[i]);
+        dp[i][1]=0;
+	}
+    for(t=2;t<=n;t++){
+        for(i=1;i<=n;i++){
+            dp[i][t]=Ma_x;
+            dq[i][t]=Mi_x; 
+            for(k=1;k<t;k++){
+         dp[i][t]=min(dp[i][t],dp[i][k]+dp[(i+k-1)%n+1][t-k]+sum(i,t));      dq[i][t]=max(dq[i][t],dq[i][k]+dq[(i+k-1)%n+1][t-k]+sum(i,t));
+	}
+        }
+    }
+    int mini=Ma_x;
+	int maxi=Mi_x; 
+    for(i=1;i<=n;i++){
+        mini=min(mini,dp[i][n]);  
+		maxi=max(maxi,dq[i][n]); 
+    }
+    printf("%d\n",mini); 
+    printf("%d ",maxi); 
+}
+实验2.3扩展距离
+#include <bits/stdc++.h>
+#include<algorithm>
+using namespace std;
+string a,b;
+int k;
+int value[10000][10000];
+int main(){
+	cin>>a>>b>>k;
+	int alen=a.size();
+	int blen=b.size();
+	for(int i=0;i<=alen;i++){
+		value[i][0]=k*i;
+	}
+	for(int i=1;i<=blen;i++){
+		value[0][i]=k*i;
+	}
+	for(int i=1;i<=alen;i++){
+		for(int j=1;j<=blen;j++){
+	value[i][j]=min(min(value[i-1][j]+k,value[i][j-1]+k),value[i-1][j-1]+abs(a[i-1]-b[j-1]));
+		}
+	}
+	cout<<value[alen][blen];
+}
+实验2.4双胞胎探宝
+#include<bits/stdc++.h>
+#include<iostream>
+using namespace std;
+int n;
+int m[15][15];
+int f[25][15][15];
+int main() {
+    cin >> n;
+    int x, y, z;
+    while (cin >> x >> y >> z) {
+        if (!x && !y && !z) break;
+        m[x][y] = z;
+    }
+    for (int k = 2; k <= 2 * n; k ++ ) 
+    for (int i = 1; i <= n; i ++ )
+    for (int j = 1; j <= n; j ++ ) {
+    int a = k - i, b = k - j;
+    if (a >= 1 && a <= n && b >= 1 && b <= n) {
+    int t = max(max(f[k - 1][i - 1][j], f[k - 1][i][j]), max(f[k - 1][i][j - 1], f[k - 1][i - 1][j - 1]));
+       if (i == j) f[k][i][j] = t + m[i][a];
+       else f[k][i][j] = t + m[i][a] + m[j][b];
+                }
+            }
+    cout << f[2 * n][n][n];
+    return 0;
+}
+实验2.5双胞胎做题
+#include <bits/stdc++.h>
+using namespace std;
+int f[1000][1000];
+int a[10000],b[10000];
+int n,sum,result=9999;
+
+void resulta(){
+	memset(f, 0, sizeof(f)); 
+	for(int i=1;i<=n;i++){
+		
+		for(int j=0;j<=sum;j++){
+			if(j<a[i]){
+				f[i][j]=f[i-1][j]+b[i];
+			}
+			else{
+				f[i][j]=min(f[i-1][j-a[i]],f[i-1][j]+b[i]);
+			}
+		}
+	}
+	for(int j=0;j<=sum;j++)       
+	{
+		int t=max(f[n][j],j);    
+		if(result>t)
+			result=t; 
+	}
+}
+
+int main(){
+	cin>>n;
+	for(int i=1;i<=n;i++){
+		cin>>a[i];
+		sum=sum+a[i];
+	}
+	for(int i=1;i<=n;i++){
+		cin>>b[i];
+	}
+	resulta();
+	cout<<result;
+} 
+实验2.6少打比赛多训练
+#include <bits/stdc++.h>
+
+using namespace std;
+typedef long long ll;
+ll dp[100002],n,m,s[100002]={0};
+map<ll,int>T;
+struct node{
+    ll u;ll t;
+}a[100002];
+int main()
+{
+    cin>>n>>m;
+    for(int i=0;i<m;i++){
+        scanf("%lld%lld",&a[i].u,&a[i].t);
+        T[a[i].u*100000+(s[a[i].u]++)]=a[i].t;//记录以a[i].u开始的每一个维持时间
+    }
+    memset(dp,0x3f,sizeof(dp));
+    dp[n+1]=0;
+    for(int i=n;i>=1;i--){
+        if(s[i]==0){dp[i]=dp[i+1];}
+        else{
+        for(int j=0;j<s[i];j++)
+                dp[i]=min(dp[i+T[i*100000+j]]+T[i*100000+j],dp[i]);
+        }
+    }
+    cout<<dp[1]<<endl;
+    return 0;
+}
+实验3.1教室安排问题
+#include<bits/stdc++.h>
+#include <iostream>
+#include <algorithm>
+
+using namespace std;
+struct huichang
+{
+	int s;
+	int f;
+};
+huichang t[10000+10];
+int n;
+bool cmp(huichang a, huichang b) {
+	return a.s < b.s;
+}
+int meeting() {
+	int count_chang = 0;
+	int *a = new int[n + 1];	//该数组于保存每个会场的结束时间
+	for (int i = 0; i < n+1; i++) {
+		a[i] = 0;
+	}
+	for (int i = 1; i <= n; i++) {
+	for (int j = 1; j <= n; j++) {
+	if (a[j] <= t[i].s) {
+	a[j] = t[i].f;
+	if (j >= count_chang + 1)					count_chang++;
+	break;
+	}
+	}
+	}
+	delete[]a;
+	return count_chang;
+}
+int main() {
+cin >> n;
+for (int i = 1; i <= n; i++) {
+cin >> t[i].s >> t[i].f;
+}
+sort(t+1,t+n+1,cmp);
+cout << meeting() << endl;
+return 0;
+}
+实验3.2磁带最优存储问题
+#include <bits/stdc++.h>
+#include<iostream>
+#include<queue>
+#include<fstream>
+using namespace std;
+struct  Tape
+{
+int len;  //长度
+int pr;  //概率    
+double tr;
+friend bool operator< (Tape a,Tape b) {return a.tr>b.tr;}
+Tape(int l=0,int p=0):len(l),pr(p),tr(l*p){};    
+};
+priority_queue<Tape>tape;
+double minTimeCost(int n,priority_queue<Tape>&tape,long long sum){
+    double res=0,acc=0;
+    while(!tape.empty())
+    {
+        acc+=tape.top().tr/sum;//计算tr累加值
+        res+=acc;
+        tape.pop();
+    }
+    return res;
+}
+int main()
+{
+    int n,a,b;
+    long long sum=0;
+    cin>>n;
+    for(int i = 0;i<n;i++){
+        cin>>a>>b;
+        tape.push(Tape(a,b));
+        sum+=b;//计算概率和
+    }
+    double res=minTimeCost(n,tape,sum);
+    cout<<res;
+    return 0;
+}    }
+    double res=minTimeCost(n,tape,sum);
+    cout<<res;
+    return 0;
+}
+作业2.1烟台基地宿舍分配
+#define _CRT_SECURE_NO_WARNINGS 1
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+typedef pair<int, int> PII;
+const int N = 1e5 + 10;
+ll n;
+ll dp[110][110];
+int main()
+{
+    ios::sync_with_stdio(0);
+    cin.tie(0);
+    cin >> n;
+    memset(dp, 0, sizeof dp);
+    dp[1][1] = 1;// i个人分j个宿舍
+    for (int i = 0; i <= n; i++) {
+        dp[i][0] = 0;
+        dp[i][1] = 1;
+        dp[0][i] = 0;
+    }
+    ll sum = 0;
+    for (int i = 1; i <= n; i++) {
+    for (int j = 1; j <= i; j++) {
+    if(dp[i][j]==0) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] * j;
+        }
+    }
+    for (int i = 1; i <= n; i++) {
+        sum += dp[n][i];
+    }
+    cout << sum;
+    return 0; //17
+}
+作业2.2烟台基地宿舍分配2
+#define _CRT_SECURE_NO_WARNINGS 1
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+typedef pair<int, int> PII;
+const int N = 1e5 + 10;
+ll n,m;
+ll dp[110][110];
+int main()
+{
+    ios::sync_with_stdio(0);
+    cin.tie(0);
+    cin >> n>>m;
+    memset(dp, 0, sizeof dp);
+    dp[1][1] = 1;// i个人分j个宿舍
+    for (int i = 0; i <= n; i++) {
+        dp[i][0] = 0;
+        dp[i][1] = 1;
+        dp[0][i] = 0;
+    }
+    ll sum = 0;
+    for (int i = 1; i <= n; i++) {
+        for (int j = 1; j <= i; j++) {
+            if(dp[i][j]==0) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] * j;
+     }
+    }
+    cout << dp[n][m];
+    return 0; 
+}
+作业3.1金字塔寻宝
+#include <bits/stdc++.h>
+#include<iostream>
+using namespace std;
+const int N=105;
+int n,*num,*lastNum,ary1[N],ary2[N];
+void dp()
+{
+num=ary1,lastNum=ary2;
+for(int i=1;i<=n;i++){
+for(int j=1;j<=i;j++){
+cin>>num[j];
+num[j]=max(lastNum[j],lastNum[j-1])+num[j];
+}
+swap(num,lastNum);
+}
+}
+int maxValue()
+{
+	int max=lastNum[1];
+	for(int i=2;i<=n;i++)
+		if(lastNum[i]>max) max=lastNum[i];
+	return max;
+}
+
+int main()
+{
+	cin>>n;dp();
+	cout<<maxValue()<<endl;
+	return 0;
+}
+作业3.2数字游戏
+#include <bits/stdc++.h>
+using namespace std;
+void getnumber(string s);
+void getresult();
+const int N=1010;
+int f[N][N],num[N][N],a[N];
+int d[N][N];
+int ans[N];
+int cnt;
+int n,k;
+int main(){
+string s;
+int c,e;
+cin>>n>>k;
+cin>>s;
+k--;
+getnumber(s);
+getresult();
+cout<<f[n][k];
+return 0;
+}
+void getnumber(string s){
+for(int i=1;i<=s.size();i++){
+a[i]=s[i-1]-'0';
+} 
+for(int i=1;i<=n;i++){
+for(int j=i;j<=n;j++){
+num[i][j]=num[i][j-1]*10+a[j];
+}
+}
+} 
+void getresult(){
+for(int i=1;i<=n;i++){ 
+f[i][0]=num[1][i]; //处理边界
+}
+for(int i=1;i<=n;i++){ //枚举处理多少个数字
+for(int j=1;j<=k;j++){ //枚举插入多少个乘号
+for(int t=1;t<i;t++){ //枚举最后乘号的乘号插入哪里（将前多少个数字划为一类）
+f[i][j]=max(f[i][j],f[t][j-1]*num[t+1][i]);
+}
+}
+}
+}
+作业4.1摸金校尉
+#include <bits/stdc++.h>
+using namespace std;
+const int N=10000;
+int servicetime[N];
+int line[N];
+int main(){
+    int n,s,sum,num;
+    cin>>n>>s;
+for(int i=0;i<n;i++){
+    cin>>servicetime[i];
+}
+sort(servicetime,servicetime+n);
+sum=servicetime[0]+s;
+num=1;
+for(int i=0;i<n;i++){
+    if(sum<servicetime[i]){
+        num+=1;
+        sum=servicetime[i]+s;
+    }
+}
+cout<<num;
+}
+作业4.2平均等待时间
+#include <bits/stdc++.h>
+using namespace std;
+const int N=10000;
+int servicetime[N];
+int line[N];
+int main(){
+    int n,s,sum;
+    cin>>n>>s;
+    for(int i=0;i<n;i++){
+        cin>>servicetime[i];
+    }
+    sort(servicetime,servicetime+n);
+    for(int i=0;i<n;i++){
+        line[i%s]+=servicetime[i];
+        sum+=line[i%s];
+    }
+    cout<<(sum*1.0/n);
+}