Codeforces Round 1013 (Div. 3)

Problem - A - Codeforces

 

解题思路：

对每个需要的数字进行计数

#include<bits/stdc++.h>
using namespace std;int main()
{int t;cin >> t;while (t--){int n;cin >> n;int two = 2;int zero = 3;int five = 1;int three = 1;int one = 1;int flag = 0;int ok = 0;for (int i = 0; i < n; i++){int now;cin >> now;if (now == 2&&two>0){two--;}if (now == 0 && zero > 0){zero--;}if (now == 5 && five > 0){five--;}if (now == 1 && one > 0){one--;}if (now == 3 && three > 0){three--;}if (two + zero + five + one+three == 0&&flag==0){flag = 1;ok = i+1;}}if (flag == 1){cout << ok << endl;}elsecout << 0 << endl;}}

Problem - B - Codeforces

解题思路：

所有人进行排序，对于大于等于x的人们，单独一队，不足的抱团看看最低能力*数量是否满足

#include <bits/stdc++.h>
using namespace std;
int n, t;
bool cmp(int a, int b)
{return a > b;
}
int main()
{int t;cin >> t;while (t--){int n, x;cin >> n >> x;vector<int>arr(n);for (int i = 0; i < n; i++){cin >> arr[i];}sort(arr.begin(), arr.end(),cmp);int alone = 0;int j = -1;for (int i = 0; i < n; i++){if (arr[i] >= x){alone++;}else{j = i;break;}}int I = 1;if (j == -1){goto there;}while (1){if (I * arr[j] >= x){alone++;I = 1;j++;}else{I++;j++;}if (j >= n){break;}}there:cout << alone << endl;}
}

Problem - C - Codeforces

解题思路：

对于偶数个数，是不可能实现的，

对于奇数个数，如果我把所有数字逆序排放，是不是就成功了？

确实是的，就这么解了，但是原理呢？

放当前数要刚好离他目标位置差x个，放一个，x++,就实现了移动一次，只有一个数在原位

来看代码：

#include <bits/stdc++.h>
using namespace std;int main() {ios::sync_with_stdio(false);cin.tie(0);int t;cin >> t;while (t--) {int n;cin >> n;if (n % 2 == 0) {cout << "-1\n";continue;}vector<int> ans(n+1);int now = n;for (int i = 0; i < n; i++){ans[i] = (now - i + n) % n;if (ans[i] == 0)ans[i] = n;cout << ans[i] << " ";}cout << endl;}return 0;
}

Problem - D - Codeforces

解决方案：

把桌子平均分给每一层，向上取整，模拟最差情况，只讨论最差情况的第一层。

每层容量m，若最差情况是放x个。那么第一层有m-x个空余，这m-x空余位就插入x中间，尽量隔开他们。

来看代码：

#include <bits/stdc++.h>
using namespace std;int main() {int t;cin >> t;while (t--) {int n, m, k;cin >> n >> m >> k;int now = (k + n- 1) / n;//每行放多少桌子int ok = m - now+1;//一行被桌子分成几部分int sum = (now + ok - 1) / ok;cout << sum << endl;}}