A
 

3 E right
10 A wrong
30 C wrong
50 B wrong
100 A wrong
200 A right
250 C wrong
300 D right
-1
7 H right
15 B wrong
30 E wrong
35 E right
80 B wrong
80 B right
100 D wrong
100 C wrong
300 C right
300 D wrong
-1
Sample Output
3 543
4 502

題意：計算罰時，輸出AC題個數和罰時。
思路：這個題AC了就將時間加起來，要是前面有WR了，一次20分鐘。 

#include <iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
#include<math.h>
#include<queue>
#include <set>
using namespace std;
typedef long long ll;
priority_queue<int> q;
set <char> s;
int main()
{
    ll ans=0,i,j,m,sum=0,n;
    char a,b[10];
    ll flag[500]={0},c[500]={0};
    while(cin>>n)
    {
        if(n==-1)
        {
            cout<<ans<<" "<<sum<<endl;
            memset(b,0,sizeof(b));
            memset(c,0,sizeof(c));
            memset(flag,0,sizeof(flag));
            sum=0;ans=0;
            continue;
        }
        cin>>a>>b;
        if(flag[a-'A']==0&&strcmp(b,"right")==0)
        {
            sum+=c[a-'A']+n;
            flag[a-'A']=1;
            ans++;
        }
        else
        if(flag[a-'A']==0)
        {
            c[a-'A']+=20;
        }
    }
    return 0;
}
 

F

Sample Input
5
JOE
BOB
ANDY
AL
ADAM
11
HOPE
ALI
BECKY
JULIE
MEGHAN
LAUREN
MORGAN
CARLI
MEGAN
ALEX
TOBIN
4
GEORGE
JOHN
PAUL
RINGO

Sample Output
DECREASING
NEITHER
INCREASING

題意：字串是升序排列，輸出INCREASING，降序DECREASING，否則NEITHER。
思路：直接sort，比較就好。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
#include<cmath>
#include<queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
typedef long long ll;
priority_queue<int> q;
set <char> s;
bool cmp(char *a,char *b)
{
    return strcmp(a,b)<0;
}
int main()
{
    ll n;
    while(cin>>n)
    {
        char s[100][100];
        char *p[100];
        ll i,j;
        for(i=0; i<n; i++)
        {
            cin>>s[i];
            p[i]=s[i];
        }
        sort(p,p+n,cmp);
        if(strcmp(p[0],s[0])==0)
        {
            ll ans=0;
            for(i=1; i<n; i++)
            {
                if(strcmp(p[i],s[i])==0)
                    ans++;
            }
            if(ans==n-1)
                cout<<"INCREASING"<<endl;
            else
                cout<<"NEITHER"<<endl;
        }
        else
        if(strcmp(p[n-1],s[0])==0)
        {
            ll ans=0;
            for(i=1,j=n-2; i<n,j>=0; i++,j--)
            {
                if(strcmp(p[j],s[i])==0)
                    ans++;
            }
            if(ans==n-1)
                cout<<"DECREASING"<<endl;
            else
                cout<<"NEITHER"<<endl;
        }
        else
            cout<<"NEITHER"<<endl;
    }
    return 0;
}
 

D

Sample Input
ABC HAPPYBIRTHDAYCACEY
ABC TRAGICBIRTHDAYCACEY
ABC HAPPYBIRTHDAY
SECRET SOMECHORESARETOUGH
SECRET SOMECHEERSARETOUGH

Sample Output
PASS
FAIL
FAIL
PASS
FAIL

題意：第一個串依次在第二個出現過，不能提前出現。
思路：將長串作為主串，比較兩個串的當前位置，相等j++，不是就繼續走（此時i++了），然後看小串的j位置是否在主串中出現過，出現直接跳出。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
#include<math.h>
#include<queue>
#include <set>
using namespace std;
typedef long long ll;
priority_queue<int> q;
set <char> s;
int main()
{
    char a[10],b[100],c[100];
    ll flag[100]={0};
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    memset(flag,0,sizeof(flag));
    while(cin>>a>>b)
    {
        ll i,j,k;
        ll n=strlen(b);
        ll nn=strlen(a);
        j=0;
        for(i=0;i<n;i++)
        {
            if(b[i]==a[j])
                j++;
            else
            {
                for(k=j;k<nn;k++)
                    if(b[i]==a[k])
                      break;
                if(k<nn)
                    break;
            }
        }
       if(i==n&&j==nn)
        cout<<"PASS"<<endl;
       else
        cout<<"FAIL"<<endl;
    }
    return 0;
}