思路：

我们发现它的比值就是原串w和b的个数比。
那我们就直接一段一段去模拟就行了。

$code$

#include<iostream>
#include<cstdio>

#define ll long long

using namespace std;

const ll MAXN = 1e5 + 10;

ll t, n;

struct node {
    
	char ch;
	ll x;
}a[MAXN];

ll gcd(ll x, ll y) {
    
	if(y == 0) return x;
	else return gcd(y, x % y);
}

ll ji(ll x, ll y, ll z) {
    
	return x * z / y;
}

ll sum(ll x, ll y) {
     // w b
	ll w = 0, b = 0, len = 0;
	for(ll i = 1; i <= n; i ++) {
    
		if(a[i].ch == 'B') {
    
			if(w * y % x == 0)
				if(b < ji(w, x, y) && b + a[i].x >= ji(w, x, y)) len ++;
			b += a[i].x;
		}
		if(a[i].ch == 'W') {
    
			if(b * x % y == 0)
				if(w < ji(b, y, x) && w + a[i].x >= ji(b, y, x)) len ++;
			w += a[i].x;
		}
	}
	return len;
}

int main() {
    
	freopen("silly.in", "r", stdin);
	freopen("silly.out", "w", stdout);
	scanf("%lld", &t);
	while(t--) {
    
		scanf("%lld", &n);
		ll b = 0, w = 0;
		for(ll i = 1; i <= n; i ++) {
    
			scanf("%lld %c", &a[i].x, &a[i].ch);
			if(a[i].ch == 'W') w += a[i].x;
			else b += a[i].x;
		}
		if(w == 0) {
    
			printf("%lld\n", b);
			continue;
		}
		if(b == 0) {
    
			printf("%lld\n", w);
			continue;
		}
		ll k = gcd(b, w);
		b /= k, w /= k;
		ll len = sum(w, b);
		printf("%lld\n", len);
	}
	return 0;
}