HDU 5432 BC - Pyramid Split(二分 + 模拟)

Pyramid Split

2
2
6 5
10 7
8
702 983 144 268 732 166 247 569
20 37 51 61 39 5 79 99

1
98这道题目可以直接二分也可以模拟通过，这里的是二分代码


#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <vector>
#include <cctype>
#include <cstdio>
#include <string>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;

#define pb push_back
#define mp make_pair
#define fillchar(a, x) memset(a, x, sizeof(a))
#define copy(a, b) memcpy(a, b, sizeof(a))

typedef long long LL;
typedef pair<int, int > PII;
typedef unsigned long long uLL;
template<typename T>
void print(T* p, T* q, string Gap = " ") {
int d = p < q ? 1 : -1;
while(p != q) {
cout << *p;
p += d;
if(p != q) cout << Gap;
}
cout << endl;
}
template<typename T>
void print(const T &a, string bes = "") {
int len = bes.length();
if(len >= 2)cout << bes[0] << a << bes[1] << endl;
else cout << a << endl;
}

const int INF = 0x3f3f3f3f;
const int MAXM = 1e5;
const int MAXN = 1e4 + 5;
const double eps = 1e-4;
int T, n, A[MAXN], B[MAXN],MAXH;

bool C(double m) {
double Mm = 0, M = 0;
for(int i = 0; i < n; i ++) {
if(m > A[i]) {
M += B[i] * B[i] * A[i] * 1.0 / 3.0;
}
else {
double Bi = (A[i] - m) * 1.0 / A[i];
M += B[i] * B[i] * A[i] * 1.0 / 3.0;
Mm += B[i] * Bi * B[i] * Bi * (A[i] - m) / 3.0;
}
}
if(M >= 2.0 * Mm) return true;
return false;
}

int main() {
cin >> T;
while(T --) {
cin >> n;
MAXH = 0;
for(int i = 0; i < n; i ++) {
cin >> A[i];
MAXH = max(MAXH, A[i]);
}
for(int i = 0; i < n; i ++) {
cin >> B[i];
}
double lb = 0, ub = MAXH;
while(ub - lb > eps) {
double mid = (ub + lb) / 2.0;
if(C(mid)) ub = mid;
else lb = mid;
}
printf("%d\n", (int)ub);
}

return 0;
}