54 lines
1.2 KiB
C++
54 lines
1.2 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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int a[1001] = {0};
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int L, n, m;
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//在以mid为最短的跳跃距离时，有多少块岩石符合要求
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int judge(int mid) {
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int cnt = 0, k = 0; //k为探测是否需要移除的点
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for (int i = 1; i <= n; i++) {
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if (a[i] - k <= mid) {
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cnt++;
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} else {
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k = a[i];
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}
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}
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return cnt;
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}
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int main() {
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//输入+输出重定向
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freopen("../1310.txt", "r", stdin);
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cin >> L >> n >> m;
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for (int i = 1; i <= n; i++) {
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cin >> a[i];
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}
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//最后的收尾岩石
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n++;
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a[n] = L;
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//二分法的起始值
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int l = 0, r = L;
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//二分法的模板
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while (l <= r) {
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int mid = (l + r) / 2;
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//如果使用这个mid，那么需要移除多少个岩石
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int x = judge(mid);
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//如果大于m, 表示意味着mid取大了，需要缩减
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if (x > m) {
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r = mid - 1;
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} else { //mid值取小了，需要增大
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l = mid + 1;
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}
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}
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//l最终是结果
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cout << l << endl;
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//关闭文件
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fclose(stdin);
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return 0;
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}
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