cjl/abc.txt

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);
}