Bnu49101:rating calculation

Whalyzh recently participated in a competition called BC (BNU Coder), which is divided into N-Fields. Participating in each account has a rating, the initial rating value is 1500, after each game the system will evaluate the participant's account, if the evaluation is good, then the account rating increase 100 points, but not more than 3299, if the evaluation of bad, The account rating down 100 points, but not less than 0 points. In order to avoid the lower the rating, Whalyzh registered two accounts, each game he will only use rating low account (if rating like a casual use). Now, he wants to know how much of the account rating is rating high after n games. Input

There are multiple sets of input data.

The first line enters an integer T (t≤10) that represents the number of data groups.

The first behavior of each group of data is two integer N (n≤10^3), which represents the number of matches.

The next n rows are a string good or bad for each row, representing the evaluation that Whalyzh obtained during the game. Output

Each set of data outputs a number that represents the rating value of the rating high account. Sample Input

2
2
good
good
2
good
bad

Sample Output

1600

Source13th session of Beijing Normal University program design competition finals AuthorHwq

#include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <queue&
Gt
#include <map> #include <set> #include <vector> #include <math.h> #include <algorithm>
using namespace Std; #define LS 2*i #define RS 2*i+1 #define UP (i,x,y) for (i=x;i<=y;i++) #define OFF (i,x,y) for (i=x;i>=y;i--) #define Me
M (a,x) memset (A,x,sizeof (a)) #define W (a) while (a) #define LL long Long const double PI = ACOs (-1.0);
#define N #define MOD 19999997 const int INF = 0X3F3F3F3F;
#define EXP 1e-8 int t,n;

Char str[100];
    int main () {int s[2];
    scanf ("%d", &t);
        W (t--) {scanf ("%d", &n);
        S[0] = s[1] = 1500;
            W (n--) {scanf ("%s", str);
            if (str[0]== ' G ') {s[0]+=100;
            } else {s[0]-=100;
            } sort (s,s+2);
          s[0]=s[0]>3299?3299:s[0];  s[0]=s[0]<0?0:s[0];
            s[1]=s[1]>3299?3299:s[1];
        s[1]=s[1]<0?0:s[1];
        } sort (s,s+2);
    printf ("%d\n", s[1]); }
}