Codeforces Round #782 (Div. 2) A-D题解

A. Red Versus Blue

每一次循环都计算R和B的比例，然后输出t个R，一个B，每次输出都判断是否结束

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>#define endl '\n'using namespace std;const int N = 1e5 + 10;void mian()
{int n, a, b;cin >> n >> a >> b;while (n){int t = 1.0 * a / (b + 1) + 0.99;for (int i = 1; i <= t; i++){cout << "R";n--;if (n == 0)break;a--;}if (n == 0)break;cout << "B";n--;b--;}cout << endl;
}int main()
{int T;cin >> T;while (T--){mian();}return 0;
}

B. Bit Flipping

题目要求是可以将一位不变，可以转换一下这种情况为使一位发生变化。但是次数要求是偶数次，所以如果k为奇数，就将全部位转换一下。

然后可以发现有k次机会使一位发生变化，最后剩余的次数全部给最低位即可。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>#define endl '\n'using namespace std;const int N = 1e5 + 10;void mian()
{int n, k;string s;cin >> n >> k;cin >> s;if (k == 0){cout << s << endl;for (int i = 1; i <= n; i++){cout << "0" << ' ';}cout << endl;}else{if (k % 2){for (int i = 0; i < s.size(); i++){s[i] ^= 1;}}vector<int> ans(n);for (int i = 0; i < s.size() - 1; i++){if (k == 0)break;if (s[i] == '0'){s[i] = '1';k--;ans[i] = 1;}elseans[i] = 0;}if (k % 2){s[s.size() - 1] ^= 1;}ans[n - 1] = k;cout << s << endl;for (auto i : ans){cout << i << ' ';}cout << endl;}
}int main()
{int T;cin >> T;while (T--){mian();}return 0;
}

C. Line Empire

题意：X轴上有n个城市，自己主城在0处，可以花费b乘上主城与另一个城市之间的距离来攻占掉，也可以花费a乘两地距离来迁都。迁都只能迁到已经攻占的地方，攻占不能跳过其他城市攻占，求花费最小多少。

如果不迁都，那么以后每次攻占都要计算上这一段距离，那么可以算一下还有几个城要攻占乘上权值b，与权值a进行比较，如果比a大那么就要迁都。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>#define endl '\n'using namespace std;typedef long long ll;const int N = 1e5 + 10;ll x[N];void mian()
{ll n, a, b, pos = 0;ll res = 0;cin >> n >> a >> b;for (int i = 1; i <= n; i++){cin >> x[i];}res += x[1] * b;for (int i = 2; i <= n; i++){if ((n - i + 1) * b > a){res += a * (x[i - 1] - pos);pos = x[i - 1];}res += b * (x[i] - pos);}cout << res << endl;
}int main()
{int T;cin >> T;while (T--){mian();}
}

D. Reverse Sort Sum

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>#define endl '\n'using namespace std;const int N = 1e5+10;void mian()
{int n;cin >> n;vector<int> a(n, 1);vector<int> c(n);for (int i = 0; i < n; i++){cin >> c[i];}for (int i = 0; i < n; i++){if (c[i] == 0){a[i] = 0;}else if (c[i] < n){a[c[i]] = 0;c[c[i]] += c[i];}}for (int i = 0; i < n; i++){cout << a[i] << ' ';}cout << endl;
}int main()
{int T;cin>>T;while(T--){mian();}return 0;
}

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