uva1450

原创于 2022-07-19 21:01:40 发布 · 123 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++

练习题目专栏收录该内容

80 篇文章

订阅专栏

该程序解决了一个数学问题，通过动态规划找到数组中需要移除的最少元素数，使得两数组元素之和不超过特定值。算法使用了二分查找优化了复杂度。

#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int maxn = 5005;
int A[maxn], B[maxn], n;

bool check(int v) {
	int ca = 0, cb = 0, remain = 0,a,b;
	for (int i = 1; i <= n; i++) {
		ca += A[i];
		cb += B[i];

		a = max(ca - v, 0);
		b = max(cb - v, 0);
		if (a + b > remain) return false;

		if (ca > 0 && cb > 0 && ca + cb > remain) remain++;
		else if (ca > 0 && cb == 0) ca--;
		else if (ca == 0 && cb > 0)cb--;
	}
	return true;
}
int main()
{
	int t;
	cin >> t;
	while (t--) {
		cin >> n;
		for (int i = 1; i <= n; i++) {
			cin >> A[i] >> B[i];
		}
		int l = 1, r = maxn * 20;
		while (l < r) {
			int mid = (l + r) >> 1;
			if (check(mid)) {
				r = mid;
			}
			else l = mid + 1;
		}
		cout << l - 1 << endl;
	}
	return 0;
}