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;
                else
                    l = 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;
}