C. Diluc and Kaeya
【思路】
记录前缀和，从1遍历到n，当前的答案等于当前 D的数量/K的数量出现的次数
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll maxn=2e5+7;
map<double,ll>mp;
int main()
{
ll t;
cin>>t;
while(t--)
{
mp.clear();
ll sum1=0;
ll sum2=0;
ll n;
cin>>n;
string s;
cin>>s;
for(ll i=0;i<s.size();i++)
{
if(s[i]==‘D‘)
{
sum1++;
}
else sum2++;
if(sum1==0)
{
cout<<sum2<<" ";
}
else if(sum2==0)
{
cout<<sum1<<" ";
}
else
{
double k=sum1*1.0/sum2;
mp[k]++;
cout<<mp[k]<<" ";
}
}
cout<<endl;
}
return 0;
}