一、algorithm
1、sort
Input：每组测试数据有两行，第一行有两个数n,m(0<n,m<1000000)，第二行包含n个各不相同，且都处于区间[-500000,500000]的整数。
Output：对每组测试数据按从大到小的顺序输出前m大的数。
Sample Input：
5 3 3 -35 92 213 -644
Sample Output：
213 92 3
代码：
#include<iostream> #include<algorithm> #include<cstdio> using namespace std; bool cmp(int a,int b){ return a>b; } int main(){ int num1,num2; while(cin>>num1>>num2){ int a[100]; for(int i=0;i<num1;i++){ scanf("%d",&a[i]); } sort(a,a+num1,cmp); for(int i=0;i<num2;i++){ cout<<a[i]<<" "; } cout<<endl; } system("pause"); return 0; }
问题2：
One thing they need to do is blowing the balloons.
Before sitting down and starting the competition, you have just passed by the room where the boys are blowing the balloons. And you have found that the number of balloons of different colors are strictly different.
After thinking about the volunteer boys' sincere facial expressions, you noticed that, the problem with more balloon numbers are sure to be easier to solve.
Now, you have recalled how many balloons are there of each color.
Please output the solving order you need to choose in order to finish the problems from easy to hard.
You should print the colors to represent the problems.
Input：The first line is an integer T which indicates the case number.
And as for each case, the first line is an integer n, which is the number of problems.
Then there are n lines followed, with a string and an integer in each line, in the i-th line, the string means the color of ballon for the i-th problem, and the integer means the ballon numbers.
It is guaranteed that:
T is about 100.
1≤n≤10.
1≤ string length ≤10.
1≤ bolloon numbers ≤83.(there are 83 teams :p)
For any two problems, their corresponding colors are different.
For any two kinds of balloons, their numbers are different.
Output：For each case, you need to output a single line.
There should be n strings in the line representing the solving order you choose.
Please make sure that there is only a blank between every two strings, and there is no extra blank.
Sample Input：
3 3 red 1 green 2 yellow 3 1 blue 83 2 red 2 white 1
Sample Output：
yellow green red blue red white
代码：
#include<iostream> #include<algorithm> #include<cstdio> #include<string> using namespace std; struct ss{ string col; int val; }num[555]; bool cmp(ss a,ss b){ //return a.val>b.val; if(a.val == b.val){ return a.col>b.col; }else{ return a.val>b.val; } } int main(){ int t; cin>>t; getchar(); while(t--){ int n; cin>>n; for(int i=0;i<n;i++){ cin>>num[i].col; cin>>num[i].val; } sort(num,num+n,cmp); for(int i=0;i<n-1;i++){ cout<<num[i].col<<" "; } cout<<num[n-1].col<<endl; } system("pause"); return 0; }
问题3：插入排序
代码：
#include <cstdio> #include <iostream> using namespace std; int main(){ int temp[100]; for(int i=1;i<=10;i++){ cin>>temp[i]; } for(int i=2;i<=10;i++){ int t = temp[i]; int j = i; while(j>1&&t>temp[j-1]){ temp[j] = temp[j-1]; j--; } temp[j] = t; } for(int i=1;i<=10;i++){ cout<<temp[i]<<" "; } cout<<endl; return 0; }
2、二分（有序序列）
eg:{1,2,2,3,3,3,5,5,5,5}
1
5 3
1 4 6 8 10
0 5
6 10
7 100000
#include<iostream> #include<algorithm> #include<cstdio> #include<cstdlib> using namespace std; int main(){ int t; scanf("%d",&t); for(int j=1;j<=t;j++){ printf("Case %d:\n",j); int n,m; scanf("%d%d",&n,&m); int temp[100010]; for(int i=0;i<n;i++){ scanf("%d",&temp[i]); } while(m--){ int q1,q2; scanf("%d%d",&q1,&q2); int begin = (int)(lower_bound(temp,temp+n,q1)-temp); int end = (int)(upper_bound(temp,temp+n,q2)-temp); printf("%d\n",end-begin); } } return 0; }
3、归并排序与逆序对个数
6 4 11 8Sample Output
1 2 3 5 6 4 1 2 3 4 5 6 7 9 8 11 10
#include<iostream> #include<algorithm> #include<cstdio> using namespace std; int main(){ int n,m; while(cin>>n>>m){ int a[2000]; for(int i=0;i<n;i++){ a[i] = i+1; } int p=0; do{ p++; if(p==m){ for(int i=0;i<n-1;i++){ cout<<a[i]<<" "; } cout<<a[n-1]<<endl; break; } }while(next_permutation(a,a+n)); } system("pause"); return 0; }
二、stack
问题1：
Input：The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single line with several words. There will be at most 1000 characters in a line.
Output：For each test case, you should output the text which is processed.
Sample Input：
3 olleh !dlrow m'I morf .udh I ekil .mca
Sample Output：
hello world! I'm from hdu. I like acm.
代码：
#include<iostream> #include<algorithm> #include<stack> #include<string> #include<cstdio> using namespace std; int main(){ string s; int t; cin>>t; getchar(); while(t--){ getline(cin,s); int len = (int)s.size(); stack<char>st; for(int i =0;i<len;i++){ if(s[i] != ' '){ st.push(s[i]); } if(s[i] == ' '||i == len-1){ while(!st.empty()){ printf("%c",st.top());//获得栈顶元素 st.pop();//弹出栈顶元素 } if(s[i] == ' '){ cout<<" "; } } } cout<<endl; } system("pause"); return 0; }
问题2：
Input：测试输入包含若干测试用例，每个测试用例占一行，每行不超过200个字符，整数和运算符之间用一个空格分隔。没有非法表达式。当一行中只有0时输入结束，相应的结果不要输出。
Output：对每个测试用例输出1行，即该表达式的值，精确到小数点后2位。
Sample Input：
1 + 2 4 + 2 * 5 - 7 / 11 0
Sample Output：
3.00 13.36
代码1（复杂，适用于所有）：
#include<iostream> #include<algorithm> #include<queue> #include<stack> #include<map> #include<cstring> #include<cstdio> #include<cstdlib> using namespace std; //cout<<"AAAA"<<endl; struct node{ double num; char op; bool flag; }; stack<node> s; queue<node> q; map<char,int> op; void change(string str); double cal(); string str; int main(){ op['+'] = op['-'] = 1; op['*'] = op['/'] = 2; while(getline(cin,str),str!="0"){ string::iterator it; for(it=str.end();it!=str.begin();it--){ if(*it == ' '){ str.erase(it); } } while(!s.empty()){ s.pop(); } change(str); printf("%0.2f\n",cal()); } system("pause"); return 0; } void change(string str){//前缀变后缀 node temp; for(int i=0;i<(int)str.size();){ if(str[i]=='('){ temp.flag = false; temp.op = str[i]; s.push(temp); i++; }else if(str[i]==')'){ while(!s.empty()&&s.top().op!='('){ q.push(s.top()); s.pop(); } s.pop(); i++; }else if(str[i]>='0'&&str[i]<='9'){ temp.flag = true; temp.num = str[i] - '0'; i++; while(i<(int)str.size()&&str[i]>='0'&&str[i]<='9'){ temp.num = temp.num*10+(str[i]-'0'); i++; } q.push(temp); }else{ temp.flag = false; while(!s.empty()&&op[str[i]]<=op[s.top().op]){ q.push(s.top()); s.pop(); } temp.op = str[i]; s.push(temp); i++; } } while(!s.empty()){ q.push(s.top()); s.pop(); } } double cal(){//计算后缀表达式:1 1 + double temp1,temp2; node cur,temp; while(!q.empty()){ cur = q.front(); q.pop(); if(cur.flag==true){ s.push(cur); }else{ temp2 = s.top().num; s.pop(); temp1 = s.top().num; s.pop(); temp.flag = true; if(cur.op == '+'){ temp.num = temp1+temp2; }else if(cur.op == '-'){ temp.num = temp1-temp2; }else if(cur.op == '*'){ temp.num = temp1*temp2; }else{ temp.num = temp1/temp2; } s.push(temp); } } return s.top().num; }
代码2（适用于带空格但没有括号的表达式）：
#include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<stack> using namespace std; int main(){ double num; while(cin>>num){ stack<double> st; char ch = getchar(); if(ch!=' '&&num==0){ break; } st.push(num); double n; char c,s; while(scanf("%c %lf%c",&c,&n,&s)!=EOF){ if(c=='+'){ st.push(n); }else if(c=='-'){ st.push(-1*n); }else if(c=='*'){ st.top() = st.top()*n; }else if(c=='/'){ st.top() = st.top()/n; } if(s!=' '){ break; } } double sum = 0.0; while(!st.empty()){ sum+=st.top(); st.pop(); } printf("%.2f\n",sum); } return 0; }
代码3（适用于不带空格和括号的表达式）：
#include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<stack> using namespace std; int main(){ double num; while(cin>>num){ stack<double> st; if(num==0){ break; } st.push(num); double n; char c; while(scanf("%c",&c)!=EOF){ if(c=='\n'){ break; } cin>>n; if(c=='+'){ st.push(n); }else if(c=='-'){ st.push(-1*n); }else if(c=='*'){ st.top() = st.top()*n; }else if(c=='/'){ st.top() = st.top()/n; } } double sum = 0.0; while(!st.empty()){ sum+=st.top(); st.pop(); } printf("%.2f\n",sum); } return 0; }
代码3（适用于不带空格和括号的表达式且用数组模拟栈）：
#include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<stack> using namespace std; int main(){ int num; int temp[100],p=0; int sum; while(cin>>num){ sum = 0; memset(temp,0,sizeof(temp)); temp[p++] = num; int n; char c; while(scanf("%c",&c)!=EOF){ if(c=='\n'){ break; } cin>>n; if(c=='+'){ temp[p] = n; p++; }else if(c=='-'){ temp[p] = -1*n; p++; }else if(c=='*'){ p--; temp[p]= temp[p]*n; p++; }else if(c=='/'){ p--; temp[p]= temp[p]/n; p++; } } for(int i=0;i<p;i++){ sum+=temp[i]; } printf("%d\n",sum); } return 0; }
三、queue
问题：
he miss his mother very much and is very scare now.You can't image how dark the room he was put into is, so poor :(.
As a smart ACMer, you want to get ACboy out of the monster's labyrinth.But when you arrive at the gate of the maze, the monste say :" I have heard that you are very clever, but if can't solve my problems, you will die with ACboy."
The problems of the monster is shown on the wall:
Each problem's first line is a integer N(the number of commands), and a word "FIFO" or "FILO".(you are very happy because you know "FIFO" stands for "First In First Out", and "FILO" means "First In Last Out").
and the following N lines, each line is "IN M" or "OUT", (M represent a integer).
and the answer of a problem is a passowrd of a door, so if you want to rescue ACboy, answer the problem carefully!
Input:The input contains multiple test cases.
The first line has one integer,represent the number oftest cases.
And the input of each subproblem are described above.
Output:For each command "OUT", you should output a integer depend on the word is "FIFO" or "FILO", or a word "None" if you don't have any integer.
Sample Input:
4 4 FIFO IN 1 IN 2 OUT OUT 4 FILO IN 1 IN 2 OUT OUT 5 FIFO IN 1 IN 2 OUT OUT OUT 5 FILO IN 1 IN 2 OUT IN 3 OUT
Sample Output:
1 2 2 1 1 2 None 2 3
代码：
#include<iostream> #include<algorithm> #include<queue> #include<stack> #include<string> #include<cstdio> using namespace std; int main(){ int t; cin>>t; while(t--){ int n; cin>>n; string s1,s2; cin>>s1; if(s1 == "FIFO"){ queue<int> qu; while(n--){ cin>>s2; if(s2 == "IN"){ int num; cin>>num; qu.push(num); }else{ if(qu.empty()){ cout<<"None"<<endl; }else{ cout<<qu.front()<<endl; qu.pop(); } } } }else{ stack<int> st; while(n--){ cin>>s2; if(s2 == "IN"){ int num; cin>>num; st.push(num); }else{ if(st.empty()){ cout<<"None"<<endl; }else{ cout<<st.top()<<endl; st.pop(); } } } } } system("pause"); return 0; }
四、set（内部自动有序、不含重复）
问题：
球赛的规则如下：
如果A打败了B，B又打败了C，而A与C之间没有进行过比赛，那么就认定，A一定能打败C。
如果A打败了B，B又打败了C，而且，C又打败了A，那么A、B、C三者都不可能成为冠军。
根据这个规则，无需循环较量，或许就能确定冠军。你的任务就是面对一群比赛选手，在经过了若干场撕杀之后，确定是否已经实际上产生了冠军。
Input输入含有一些选手群，每群选手都以一个整数n(n<1000)开头，后跟n对选手的比赛结果，比赛结果以一对选手名字（中间隔一空格）表示，前者战胜后者。如果n为0，则表示输入结束。
Output对于每个选手群，若你判断出产生了冠军，则在一行中输出“Yes”，否则在一行中输出“No”。
Sample Input
3 Alice Bob Smith John Alice Smith 5 a c c d d e b e a d 0
Sample Output
Yes No
代码：
#include<iostream> #include<algorithm> #include<set> #include<string> #include<cstdio> using namespace std; int main(){ int t; while(cin>>t){ if(t == 0){ break; } set<string> st1,st2; string s1,s2; while(t--){ cin>>s1>>s2; st1.insert(s1); st1.insert(s2); st2.insert(s2); } if(st1.size()-st2.size() != 1){ cout<<"No"<<endl; }else{ cout<<"Yes"<<endl; } } system("pause"); return 0; }
五、迭代器
问题：
呵呵，很简单吧？
Input每组输入数据占1行,每行数据的开始是2个整数n(0<=n<=100)和m(0<=m<=100),分别表示集合A和集合B的元素个数，然后紧跟着n+m个元素，前面n个元素属于集合A，其余的属于集合B. 每个元素为不超出int范围的整数,元素之间有一个空格隔开.
如果n=0并且m=0表示输入的结束，不做处理。
Output针对每组数据输出一行数据,表示A-B的结果,如果结果为空集合，则输出“NULL”,否则从小到大输出结果,为了简化问题，每个元素后面跟一个空格.
Sample Input
3 3 1 2 3 1 4 7 3 7 2 5 8 2 3 4 5 6 7 8 0 0
Sample Output
2 3 NULL
代码：
#include<iostream> #include<algorithm> #include<set> #include<string> #include<cstdio> using namespace std; int main(){ int num1,num2; while(cin>>num1>>num2){ if(num1 == num2&&num1 == 0){ break; } int num; set<int> st; for(int i=0;i<num1;i++){ cin>>num; st.insert(num); } for(int i=0;i<num2;i++){ cin>>num; if(st.find(num) != st.end()){ st.erase(num); } } if(st.size() == 0){ cout<<"NULL"<<endl; continue; } set<int>::iterator it; for(it=st.begin();it!=st.end();it++){ cout<<*it<<" "; } cout<<endl; } system("pause"); return 0; }
六、map(键值对)
问题：
InputOne line contians a number n ( n<=10000),stands for the number of shops.
Then n lines ,each line contains a string (the length is short than 31 and only contains lowercase letters and capital letters.)stands for the name of the shop.
Then a line contians a number m (1<=m<=50),stands for the days .
Then m parts , every parts contians n lines , each line contians a number s and a string p ,stands for this day ,the shop p 's price has increased s.
OutputContains m lines ,In the ith line print a number of the shop "memory" 's rank after the ith day. We define the rank as :If there are t shops' price is higher than the "memory" , than its rank is t+1.Sample Input
3 memory kfc wind 2 49 memory 49 kfc 48 wind 80 kfc 85 wind 83 memory
Sample Output
1 2
代码：
#include<iostream> #include<algorithm> #include<map> #include<string> #include<cstdio> using namespace std; int main(){ int t; while(cin>>t){ for(int i=0;i<t;i++){ string s; cin>>s; } int n; cin>>n; map<string,int> shop; while(n--){ for(int i=0;i<t;i++){ string name; int price; cin>>price; cin>>name; shop[name]+=price; } map<string,int>::iterator it; int mas = 1; for(it=shop.begin();it!=shop.end();it++){ if(it->second>shop["memory"]){ mas++; } } cout<<mas<<endl; } } system("pause"); return 0; }
七、priority_queue(优先队列)
问题：
不过经过细心的0068的观察，他发现了医院里排队还是有讲究的。0068所去的医院有三个医生（汗，这么少）同时看病。而看病的人病情有轻重，所以不能根据简单的先来先服务的原则。所以医院对每种病情规定了10种不同的优先级。级别为10的优先权最高，级别为1的优先权最低。医生在看病时，则会在他的队伍里面选择一个优先权最高的人进行诊治。如果遇到两个优先权一样的病人的话，则选择最早来排队的病人。
现在就请你帮助医院模拟这个看病过程。
Input输入数据包含多组测试，请处理到文件结束。
每组数据第一行有一个正整数N(0<N<2000)表示发生事件的数目。
接下来有N行分别表示发生的事件。
一共有两种事件：
1:"IN A B",表示有一个拥有优先级B的病人要求医生A诊治。(0<A<=3,0<B<=10)
2:"OUT A",表示医生A进行了一次诊治，诊治完毕后，病人出院。(0<A<=3)
Output对于每个"OUT A"事件，请在一行里面输出被诊治人的编号ID。如果该事件时无病人需要诊治，则输出"EMPTY"。
诊治人的编号ID的定义为：在一组测试中，"IN A B"事件发生第K次时，进来的病人ID即为K。从1开始编号。
Sample Input
7 IN 1 1 IN 1 2 OUT 1 OUT 2 IN 2 1 OUT 2 OUT 1 2 IN 1 1 OUT 1
Sample Output
2 EMPTY 3 1 1
代码：
#include<iostream> #include<algorithm> #include<queue> #include<string> #include<cstdio> using namespace std; struct node{ int value,id; bool operator<(const node &b)const{ if(value == b.value){ return id>b.id; }else{ return value<b.value; } } }; int main(){ int t; while(cin>>t){ node pep; priority_queue<node> q[10]; int time = 1; string s; while(t--){ cin>>s; if(s == "IN"){ int a; cin>>a>>pep.value; pep.id = time; time++; q[a-1].push(pep); }else{ int a; cin>>a; if(q[a-1].empty()){ cout<<"EMPTY"<<endl; }else{ cout<<q[a-1].top().id<<endl; q[a-1].pop(); } } } } system("pause"); return 0; }
八、list
问题：
Input本题有多个测试数据组，第一行为组数N，接着为N行新兵人数，新兵人数不超过5000。
Output共有N行，分别对应输入的新兵人数，每行输出剩下的新兵最初的编号，编号之间有一个空格。
Sample Input
2 20 40
Sample Output
1 7 19 1 19 37
代码1：
#include<iostream> #include<algorithm> #include<queue> #include<string> #include<cstdio> #include<stack> #include<list> #include<cmath> #include<cstring>// #include<cstdlib>// using namespace std; int main(){ int t,n; cin>>t; while(t--){ cin>>n; int k = 2; list<int> blist; list<int>::iterator it; for(int i=1;i<=n;i++){ blist.push_back(i); } while(blist.size()>3){ int num =1; for(it=blist.begin();it!=blist.end();){ if(num++%k == 0){ it = blist.erase(it); }else{ it++; } } if(k == 2){ k = 3; continue; }else{ k = 2; continue; } //k==2?k=3:k=2; } for(it=blist.begin();it!=blist.end();it++){ if(it!=blist.begin()){ cout<<" "; } cout<<*it; } cout<<endl; } system("pause"); return 0; }
代码2：
