acm 竞赛题xiaoA is a beautiful girl,and she has so many skirts
acm 竞赛题xiaoA is a beautiful girl,and she has so many skirts that she won't wear a skirt a second time before the skirt is washed.Now xiaoA wants to go shopping,how many skirts she can choose to wear.In order to distinguish these skirts,she gives out every skirt with a unique name.Input The first line of input contains a positive integer T (T
最佳回答
#include
#include
#include
#include
using namespace std;
map gg;
int cases;
int main ()
{
scanf ("%d\n",&cases);
while (cases--)
{
int n,m;
gg。clear ();
scanf ("%d %d\n",&n,&m);
char str[20];
for (int i = 0; i < n; i++)
{
cin >> str;
gg[str] = true;
}
char dw;
for (int i = 0; i < m; i++)
{
cin >> dw >> str;
if (dw == 'D') gg[str] = false;
else if (dw = 'W' ) gg[str] = true;
}
map ::iterator iter;
int count = 0;
for (iter = gg。begin (); iter != gg。end();iter++)
{
if (iter->second)
count ++;
}
printf ("%d\n",count);
}
return 0;
}
#include
#include
#include
using namespace std;
map gg;
int cases;
int main ()
{
scanf ("%d\n",&cases);
while (cases--)
{
int n,m;
gg。clear ();
scanf ("%d %d\n",&n,&m);
char str[20];
for (int i = 0; i < n; i++)
{
cin >> str;
gg[str] = true;
}
char dw;
for (int i = 0; i < m; i++)
{
cin >> dw >> str;
if (dw == 'D') gg[str] = false;
else if (dw = 'W' ) gg[str] = true;
}
map ::iterator iter;
int count = 0;
for (iter = gg。begin (); iter != gg。end();iter++)
{
if (iter->second)
count ++;
}
printf ("%d\n",count);
}
return 0;
}
最新回答共有2条回答
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2026-04-06 20:36:56靓丽的钢笔
回复#include #include #include #include using namespace std;map gg;int cases;int main (){scanf ("%d\n",&cases);while (cases--){int n,m;gg。clear ();scanf ("%d %d\n",&n,&m);char str[20];for (int i = 0; i > str;gg[str] = true;}char dw;for (int i = 0; i > dw >> str;if (dw == 'D') gg[str] = false;else if (dw = 'W' ) gg[str] = true;}map ::iterator iter;int count = 0;for (iter = gg。begin (); iter != gg。end();iter++){if (iter->second)count ++;}printf ("%d\n",count);}return 0;}
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