程序设计实习之STL专项练习
A:List
使用list来做这题， 自己写合并结果MLE,用了list自带的merge合并就可以过。。。至今未明。。。
#include
#include
#include
#include
#include
#include
using namespace std;
char cmd[20];
int id,id1,id2,num;
list *s[10000+10];
int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
int n ;
cin>>n;
for (int i = 0; i < n; ++i)
{
scanf("%s",cmd);
switch(cmd[0])
{
case 'n':
scanf("%d",&id);
s[id] = new list;
s[id]->clear();
break;
case 'a':
scanf("%d %d",&id,&num);
s[id]->push_back(num);
break;
case 'm':
scanf("%d %d", &id1,&id2);
//if (s[id1]->size() < s[id2]->size()) swap(s[id1],s[id2]);
//for (list::iterator i = s[id2]->begin(); i != s[id2]->end(); i++)
// s[id1]->push_back(*i);
s[id1]->merge(*s[id2]);
//s[id2]->clear();
break;
case 'u':
scanf("%d",&id);
s[id]->sort();
s[id]->unique();
break;
case 'o':
scanf("%d",&id);
s[id]->sort();
for (list::iterator i = s[id]->begin(); i != s[id]->end(); i++)
printf("%d ",*i);
printf("\n");
break;
default:
break;
}
}
return 0;
}
B:RPN Calculator
stack 操作后缀表达式
因为没考虑里面有负数的情况 RE的次数略多
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
multiset memory;
stack cal;
int n;
double d;
char cmd[200];
void calc(char ope)
{
double x = cal.top(); cal.pop();
double y = cal.top(); cal.pop();
double z=0;
switch (ope)
{
case '+':
z = x+y;break;
case '-':
z = y-x;break;
case '*':
z = x*y;break;
case '/':
z = y/x;break;
case '^':
z = pow(y,x);break;
default:
break;
}
cal.push(z);
}
int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
cin>>n;
for (int i = 1; i <= n ;++i)
{
scanf("%lf",&d);
memory.insert(d);
}
while (scanf("%s",cmd) != EOF)
{
if (strlen(cmd)>1)
{
sscanf(cmd, "%lf", &d);
cal.push(d);
continue;
}
switch(cmd[0])
{
case '+':
case '-':
case '*':
case '/':
case '^':
calc(cmd[0]);
break;
case '=':
printf("%e\n", cal.top());
memory.erase(memory.begin());
memory.insert(cal.top());
while (!cal.empty()) cal.pop();
break;
default:
sscanf(cmd, "%lf", &d);
cal.push(d);
break;
}
}
cout<::iterator it = memory.begin();
for (int i = 1;it != memory.end(); it++, i++)
{
printf("%e ",*it);
if (i%10==0) printf("\n");
}
return 0;
} C:Set
开两个set记录有无出现过。
#include
#include
#include
#include
#include
#include
using namespace std;
multiset * multiset_s;
set * set_s;
int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
int n,x; char cmd[10];
cin>>n;
multiset_s = new multiset;
set_s = new set;
for (int i = 0; i < n;++i)
{
scanf("%s", cmd);
switch(cmd[1])
{
case 'd':
scanf("%d",&x);
multiset_s->insert(x);
set_s->insert(x);
printf("%d\n",multiset_s->count(x));
break;
case 'e':
scanf("%d",&x);
printf("%d\n",multiset_s->count(x));
multiset_s->erase(x);
break;
case 's':
scanf("%d",&x);
printf("%d ",(set_s->find(x)==set_s->end()?0:1));
printf("%d\n",multiset_s->count(x));
break;
default:
break;
}
}
return 0;
}
D:字符串操作
#include
#include
#include
#include
#include
#include
using namespace std;
string s[50],cmd,s1,s2;
int n,N,x,l,p;
int getInt();
string getstring();
bool isDight(string s)
{
if (s.size()>=6) return false;
for (int i = 0; i '9' || s[i]<'0') return false;
return true;
}
int rfind()
{
string s1 = getstring();
int N = getInt();
int p = s[N].rfind(s1);
if (p == string::npos)
return s[N].size();
else return p;
}
int find()
{
string s1 = getstring();
int N = getInt();
int p = s[N].find(s1);
if (p == string::npos)
return s[N].size();
else return p;
}
string copy()
{
int N = getInt();
int x = getInt();
int l = getInt();
return s[N].substr(x,l);
}
string add()
{
string s1 = getstring();
string s2 = getstring();
if (isDight(s1) && isDight(s2))
{
int k = atoi(s1.c_str())+atoi(s2.c_str());
string s3="";
while (k>0) s3= char(k%10+'0')+s3,k/=10;
return s3;
}
else
return s1+s2;
}
string insert()
{
string s1 = getstring();
int N = getInt();
int x = getInt();
return s[N].insert(x,s1);
}
void reset()
{
string s1 = getstring();
int N = getInt();
s[N] = s1;
}
int getInt()
{
string temp;
cin>>temp;
if (temp == "find")
return find();
else if (temp == "rfind")
return rfind();
else
{
int num = 0;
for (int i = 0; i >temp;
if (temp == "copy")
return copy();
else if (temp == "add")
return add();
else
return temp;
}
int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
cin>>n;
for (int i= 1; i <= n;++i)
cin>>s[i];
while (true)
{
cin>>cmd;
switch(cmd[0])
{
case 'o':
exit(0);
break;
case 'c':
s[0] = copy();
break;
case 'a':
s[0] = add();
case 'f':
s[0] = find();
break;
case 'r':
if (cmd == "rfind")
{
s[0] = rfind();
} else
reset();
break;
case 'i':
s[0] = insert();
break;
case 'p':
if (cmd.size() == 5)
cout << s[getInt()]< E:热血格斗场
#include
#include
#include
#include
#include
using namespace std;
int n,id,value;
int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
cin>>n;
set< pair > Person;
Person.insert(make_pair(1000000000,1));
set< pair >::iterator it, temp;
for (int i = 2; i <= n+1; ++i)
{
scanf("%d %d",&id,&value);
it= Person.lower_bound(make_pair(value,id));
if (it != Person.end())
temp = it;
if (it != Person.begin())
{
it --;
if ( value - it->first <= temp->first - value)
temp = it;
}
Person.insert(make_pair(value,id));
printf("%d %d\n",id,temp->second);
}
return 0;
} F:冷血格斗场
#include
#include
#include
#include
#include
using namespace std;
int n,id,value;
int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
cin>>n;
set< pair > Person;
Person.insert(make_pair(1000000000,1));
set< pair >::iterator it, temp;
for (int i = 2; i <= n+1; ++i)
{
scanf("%d %d",&id,&value);
it= Person.lower_bound(make_pair(value,-1));
if (it != Person.end())
temp = it;
if (it != Person.begin())
{
it --;
it= Person.lower_bound(make_pair(it->first,-1));
if ( value - it->first < temp->first - value || value - it->first == temp->first - value
&& it->second < temp->second)
temp = it;
}
Person.insert(make_pair(value,id));
printf("%d %d\n",id,temp->second);
}
return 0;
} G:priority queue练习题
这题用set 来做应该是很好做的 删除begin() 和 end() 的元素
用堆来做同时删除最小和最大优点困难，我的做法是用两个堆，小根堆和大根堆，然后用同一个bool数组记录有无被删除，在小根堆中被删除，即使大根堆中
来保留的话，表示为false后的元素，再次从堆中被弹出的时候，视为无效即可。
#include
#include
#include
#include
#include
#include
using namespace std;
const int N = 10000000;
int cnt,prime[N+10];
bool flag[N+10],inQueue[10000];
struct Node
{
int PrimeNumber;
int value;
int id;
}v[300+10];
void getPrime()
{
cnt=0;
memset(flag,false,sizeof(flag));
for(int i=2;i=prime[i] && i < N)
{
if (p%prime[i]==0)
{
total++;
while (p%prime[i]==0) p/=prime[i],number ++;
}
i++;
}
if (number == 1) total =0;
return total;
}
struct cmpMax
{
bool operator ()(const Node & a,const Node & b)
{
return (a.PrimeNumber < b.PrimeNumber || a.PrimeNumber == b.PrimeNumber
&& a.value < b.value);
}
};
struct cmpMin
{
bool operator ()(const Node & a,const Node & b)
{
return (a.PrimeNumber > b.PrimeNumber || a.PrimeNumber == b.PrimeNumber
&& a.value > b.value);
}
};
int main()
{
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
int n,num;
cin>>n;
getPrime();
priority_queue, cmpMax > qMax;
priority_queue, cmpMin > qMin;
memset(inQueue, true, sizeof(inQueue));
for (int test= 0; test