Educational Codeforces Round 106 (Rated for Div. 2)（A ~ E）题解（每日训练 Day.16 ）

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Educational Codeforces Round 106 (Rated for Div. 2)（A ~ D）

A. Domino on Windowsill

Problem

Solution

签到题…画个图算一下就好

Code

// Problem: A. Domino on Windowsill
// Contest: Codeforces - Educational Codeforces Round 106 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1499/problem/A
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>using namespace std;
const int N = 50007;int n, k1, k2, w, b;void solve()
{scanf("%d%d%d%d%d", &n, &k1, &k2, &w, &b);if(k1 < k2) swap(k1, k2);if(w <= (k1 - k2) / 2 + k2 && b <= (k1 - k2) / 2 + n - k1) {puts("YES");}else puts("NO");
}int main()
{int t;scanf("%d", &t);while(t -- ) {solve();}return 0;
}

B. Binary Removals

Problem

Solution

经典排序题，显然相邻的不能同时丢掉，也就意味着如果出现两个需要丢掉的数是相邻的，那么就失败了，否则一定成功。所以我们只需要找一下两个需要丢掉的，也就是前面是两个1，后面是两个0，这样就没办法同时丢掉，输出 NO

Code

// Problem: B. Binary Removals
// Contest: Codeforces - Educational Codeforces Round 106 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1499/problem/B
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>using namespace std;const int N = 50007;int n, m, t;
string a;void solve()
{bool left = false, right = false;cin >> a;int len = a.length();int pos = -1;for(int i = 0; i < len; ++ i) {if(a[i] == '1' && a[i + 1] == '1') {left = true;pos = i + 2;break;}}for(int i = pos; i < len; ++ i) {if(a[i] == '0' && a[i + 1] == '0') {right = true;break;}}if(left && right) {puts("NO");}else puts("YES");
}int main()
{scanf("%d", &t);while(t -- ) {solve();}return 0;
}

C. Minimum Grid Path

Problem

Solution

显然第一印象就是我们对于 cic_ici 小的就多走一点，cic_ici 大的就少走一点，发现少走一点，也就至少走一步，如果我们当前的 cic_ici 是最小的，那么显然直接走到头也就是走完这个方向的所有步数是最优的。有了这个印象以后，我们分析题意，发现每次选择方向必须是不同的方向，所以我们走的时候必须上，右，上，右…这样交替地行走，所以我们直接奇偶交替行走，分别选择不同的 cic_ici ，因为往右走和往上走都是一样的，都是走 nnn 步，一共走 2×n2\times n2×n 步，所以向上向右，谁奇谁偶是无所谓的，我们O(n)O(n)O(n)贪心地走一遍即可。

Code

// Problem: C. Minimum Grid Path
// Contest: Codeforces - Educational Codeforces Round 106 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1499/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>using namespace std;
#define int long long
const int N = 500007, INF = 2e9;int t, n, m;
int a[N];void solve()
{scanf("%lld", &n);for(int i = 1; i <= n; ++ i) scanf("%lld", &a[i]);int even = INF, odd = INF, cnt_even = 0, cnt_odd = 0;int ans_even = 0, ans_odd = 0;int ans = (a[1] + a[2]) * n;for(int i = 1; i <= n; ++ i) {if(i & 1) {ans_odd += a[i];cnt_odd ++ ;odd = min(odd, a[i]);}else {ans_even += a[i];cnt_even ++ ;even = min(even, a[i]);}ans = min(ans, ans_even + even * (n - cnt_even) + ans_odd + odd * (n - cnt_odd));}printf("%lld\n", ans);
}signed main()
{scanf("%lld", &t);while(t -- ) {solve();}return 0;
}

D. The Number of Pairs

Problem

Solution

这题之前我写过hhh放一下当时的题解吧，当时直接秒的（那个 t=gcd⁡(a,b)t=\gcd(a,b)t=gcd(a,b) 一定是整数所以一定整除）

以及新鲜出炉的代码（虽然跟之前我写的那份代码没有任何本质上的区别）

Code

// Problem: D. The Number of Pairs
// Contest: Codeforces - Educational Codeforces Round 106 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1499/problem/D
// Memory Limit: 512 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>using namespace std;
#define int long long
const int N = 20000007;int primes[N], cnt;
int n, t;
bool vis[N];
int m[N];
int c, d, x;void init(int n)
{for(int i = 2; i <= n; ++ i) {if(vis[i] == 0) {primes[ ++ cnt] = i;m[i] = 1;}for(int j = 1; j <= cnt && i * primes[j] <= n; ++ j) {vis[i * primes[j]] = true;if(i % primes[j] == 0) {m[i * primes[j]] = m[i];break;}m[i * primes[j]] = m[i] + 1;}}
}int f(int y)
{if((y + d) % c != 0) return 0;return 1 << (m[(y + d) / c]);
}void solve()
{int ans = 0;scanf("%lld%lld%lld", &c, &d, &x);for(int i = 1; i * i <= x; ++ i) {if(x % i == 0) {ans += f(i);if(i * i != x) ans += f(x / i);}}printf("%lld\n", ans);
}signed main()
{init(N - 7);scanf("%lld", &t);while(t -- ) {solve();} return 0;
}

E. Chaotic Merge

我DP不太行，丢给我队友了，等神仙队友写完了更…

官方题解：https://codeforces.com/blog/entry/88812

F. Diameter Cuts

树形DP，有点像2020ICPC南京的M题

G. Graph Coloring

300+的标答就离谱（