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一.算法题总结

1.

 解题思路：对于这个题，我最开始想到就是二分，但是头痛的是有三个解，如果我在-100到100之间二分，那么只能得出一个解，然后我就想了一下，这个要求精度，那么0.01这么小，好像可以在0-1之间或者-1-1之间二分，然后我就觉得好像可以遍历这俩百个格子，每个格子长度为一，进行二分求解

#include <iostream>
#include <vector>
#include <algorithm>
#define ll long long
using namespace std;double a, b, c, d;
double f(double x) {return a * x * x * x + b * x * x + c * x + d;
}int main() {cin >> a >> b >> c >> d;int ans = 0;for (int i = -100; i < 100; i++) {if (ans == 3)break;double l = i;double r = i + 1;double f1 = f(l);double f2 = f(r);if (f1 == 0) {ans++;printf("%.2f ", l);continue;}if (f1 * f2 < 0) {while (r - l >= 0.001) {double mid = (l + r) / 2;if (f(mid) * f(l) <= 0)r = mid;elsel = mid;}printf("%.2f ", l);ans++;}}return 0;
}

2.

解题思路:对于这个题，我写过Section I,所以就看出了要用贪心， 但在这个贪心要怎么用呢，也没给我个具体的数，让区间的和小于某个数，这个时候，我想起了二分，用二分来猜数的大小，而这个数的大小，就是看这个区间的是否大于它，这就可以贪心了

#include <iostream>
#include <vector>
#include <algorithm>
#define ll long long
using namespace std;bool isPossible(int mid, const vector<int>& nums, int m) {int count = 1;int current_sum = 0;for (int num : nums) {if (current_sum + num <= mid) {current_sum += num;}else {current_sum = num;count++;if (count > m) {return false;}}}return true;
}int main() {int n, m;cin >> n >> m;vector<int> nums(n);int max_num = 0;int total_sum = 0;for (int i = 0; i < n; ++i) {cin >> nums[i];max_num = max(max_num, nums[i]);total_sum += nums[i];}int left = max_num;int right = total_sum;int answer = total_sum;while (left <= right) {int mid = left + (right - left) / 2;if (isPossible(mid, nums, m)) {answer = mid;right = mid - 1;}else {left = mid + 1;}}cout << answer << endl;return 0;
}

3.

解题思路：今天写的最难的一个题，其实怪自己眼挫，没注意到（尽可能让前面的人少抄），其实这个题的本质也是贪心+二分，我就是这么写的，思路和上面的题一样，但是一直只能过4个，然后看了一下别人的题解，发现漏了个条件（尽可能让前面的人少抄），有了这个以后，那就反过来在对答案抄一边就行了

#include <iostream>
#include <vector>
#include <algorithm>
#define ll long long
using namespace std;int m, k;
struct lr {ll l, r;
};
vector<lr> len;bool check(ll x, vector<ll>& nums) {ll sum = 0;int cnt = 1;vector<lr> temp;lr current; current.l = 1;for (int i = 1; i <= m; i++) {if (nums[i] > x) return false;if (sum + nums[i] <= x) {sum += nums[i];} else {current.r = i - 1;temp.push_back(current);cnt++;sum = nums[i];current.l = i;}if (cnt > k) return false;}current.r = m;temp.push_back(current);if (cnt == k) {len = temp;return true;}return cnt<=k;
}void check2(ll x, vector<ll>& nums) {ll sum = 0;int cnt = 1;vector<lr> temp;lr current; current.r = m;for (int i = m; i >= 1; i--) {if (sum + nums[i] <= x) {sum += nums[i];} else {current.l = i + 1;temp.push_back(current);cnt++;sum = nums[i];current.r = i;}}current.l = 1;temp.push_back(current);reverse(temp.begin(), temp.end());len = temp;
}int main() {cin >> m >> k;vector<ll> nums(m + 1);ll left = 0, right = 0;for (int i = 1; i <= m; i++) {cin >> nums[i];right += nums[i];if (nums[i] > left) left = nums[i];}ll ans = right;while (left <= right) {ll mid = (left + right) / 2;if (check(mid, nums)) {ans = mid;right = mid - 1;} else {left = mid + 1;}}// 确保前面的人尽可能少抄写check2(ans, nums);for (int i = 0; i < k; i++) {cout << len[i].l << " " << len[i].r << endl;}return 0;
}