枚举总的f大小作为背包容量即可，f为0的可以直接入答案，难就难在发现f的枚举有效范围在[1,sqrt(n+2)]之间，然后就是输入输出，直接用浮点或者什么双精度肯定是错的，建议用字符串模拟读入输出，不然会wa到怀疑人生。
#include "bits/stdc++.h"
using namespace std;
#define int long long
const int N = 2e4 + 100;
int f[N];
long long s[N];
long long dp[N];
inline void solve() {
int T;
cin >> T;
while (T--) {
int n;
cin >> n;
int sumf = 0;
for (int i = 1; i <= n; ++i) {
string s1;
cin >> s[i] >> s1;
f[i] = (s1[0]-'0') * 100 + (s1[2] - '0') * 10 + s1[3] - '0';
sumf += f[i];
}
while (sumf * sumf >= 10000 * (n + 2)) sumf--;
long long ma = 0;
for (int i = 0; i <= sumf; ++i) {
for (int j = 0; j <= i; ++j) {
dp[j] = 0;
}
for (int j = 1; j <= n; ++j) {
if (f[j] ==0 || f[j] > i) continue;
long long ans = s[j] * (10000 - (i - f[j]) * f[j]);
if (ans <= 0) continue;
for (int k = i; k >= f[j]; --k) {
dp[k] = max(dp[k], dp[k - f[j]] + ans);
}
}
ma = max(ma, dp[i]);
}
for (int i = 1; i <= n; ++i) {
if (f[i] == 0)
ma += s[i] * 10000;
}
string str;
int tmp = ma % 10000;
while (tmp)
{
str = char( '0' + tmp % 10 )+ str ;
tmp /= 10;
}
while (str.length() < 4)
{
str = '0'+str;
}
cout << ma /10000 << '.' << str << "00000" << endl;
}
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}