1351 I want 4444

Description

M country has only four license plate numbers, and other people want them to be 8888,6666, 3333, etc., but Mr. L is very strange, he wants 4444 ......

M's license plate number issuing system is very simple, in order, from 0000 to 9999 a total of 10000 numbers, their cars are not much, and there may be decommissioned license plate numbers at any time also abandoned.

Now we know that we have handled K licenses, K consecutive Licenses Starting from 0000, mr. l it is quite uncomfortable. For some reason, it takes M days for him to apply for a license. as an administrator, you already know about the M-day license reservation. There are a total of P booking information, there are two types of reservation:

1. If you have an individual R to make an appointment to apply for a license today, you will be asked to give them a license from 0000;

2. If a license is to be decommissioned today, leave this position empty.

3. In the same day, we always process decommission tasks first and then new applications.

Mr. l wants to know that M days later, he wants to get a license while he is not ......

Input

The first row contains an integer T, indicating that there are T groups of data.

The first row of each data group contains two integers, K and P, which are consistent with the topic description, where P <= 50.

Each row of the following P rows contains one of the two information formats:

X n l indicates that there are L applications on Day X, and N indicates the flag character.

X d r indicates that R is to be decommissioned on Day X, and D is a flag character to ensure that the data is valid.

Ensure that the input data appears before the input data in a small amount, but do not guarantee that D must appear before n.

Output

After processing the p Reservation, the number 4444 is returned, but "yes" is returned. Otherwise, "no" is returned ".

 

 

 

Simulate questions and process data as required.

#include <stdio.h>#include<string.h>main(){int a[10000];int bannumber,opnumber;int number,te;int i,j,up;char op;int day,car;int daytemp;int sum;scanf("%d",&number);for(te=1;te<=number;te++){memset(a,0,sizeof(a));scanf("%d %d",&bannumber,&opnumber);for(i=0;i<bannumber;i++)a[i]=1;daytemp=0;        sum=0;if(opnumber==0){  if(bannumber>=4444)printf("No\n");elseprintf("Yes\n");} else {for(up=0;up<opnumber;up++){scanf("%d %c %d",&day,&op,&car);     if((daytemp==0||daytemp==day)&&op=='N') {    sum+=car; daytemp=day;      } if((daytemp==0||daytemp==day)&&op=='D') {    a[car]=0; op='A';  }          if((daytemp!=day||daytemp==0)||up==opnumber-1) {   for(j=0;j<sum;j++)         for(i=0;i<10000;i++) {        if(a[i]==0){      a[i]=1;       break;} }  daytemp=0;     if(op=='N')      sum=car; else sum=0; } if((daytemp==0||daytemp==day)&&op=='D') {    a[car]=0;  } }if(a[4444]==0)printf("Yes\n");elseprintf("No\n");}}}