Educational Codeforces Round 49 (Rated for Div. 2)ABCD题解

版权声明：博主是菜鸡，但转的时候还是说一声吧 https://blog.csdn.net/qq_37666409/article/details/82191442

A

题意大概就是给出一个字符串，每个字符必须变成它的前一个或后一个（a和z只能变成1个），问你能不能变成一个回文串

直接两个指针同时从中间向两边挪，暴力模拟即可

#include <bits/stdc++.h>
using namespace std;

#define Y puts("YES")
#define N puts("NO")

const int MAXN = 200;
const int INF = 0x3f3f3f3f;

template <typename T> inline void read(T &x) {
    int c = getchar();
    bool f = false;
    for (x = 0; !isdigit(c); c = getchar()) {
        if (c == '-') {
            f = true;
        }
    }
    for (; isdigit(c); c = getchar()) {
        x = x * 10 + c - '0';
    }
    if (f) {
        x = -x;
    }
}

int T, n;
char s[MAXN];

signed main() {
	read(T);
	while(T --) {
		read(n);
		scanf("%s", s);
		int len = strlen(s);
		int mid = (len >> 1);
		int r = mid, l = mid - 1;
		bool fg = 0;
		while(r < len && l >= 0) {
			if(s[l] == 'a') {
				if(s[r] != 'a' && s[r] != 'c') {
					fg = 1;
					N;
					break;
				}
			}
			else if(s[l] == 'z') {
				if(s[r] != 'z' && s[r] != 'x') {
					fg = 1;
					N;
					break;
				}
			}
			else {
				if(s[r] - 1 == s[l] + 1) fg = 0;
				else if(s[r] + 1 == s[l] - 1) fg = 0;
				else if(s[r] == s[l]) fg = 0;
				else {
					fg = 1;
					N;
					break;
				}
			}
			l--, r++;
			//printf("!!!%d %d\n", l, r);
		}
		if(!fg) Y;
	}
	return 0;
}

B

说实话这题没太看懂，大概意思是1-n^2/2的数从左往右从上往下只填奇数行奇数列和偶数行偶数列，剩下的相反，给出(x,y)，求其对应的数。

其实看看就知道是个公式题，算是很好推的了，具体看代码

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 100100;
const int INF = 0x3f3f3f3f;

template <typename T> inline void read(T &x) {
    int c = getchar();
    bool f = false;
    for (x = 0; !isdigit(c); c = getchar()) {
        if (c == '-') {
            f = true;
        }
    }
    for (; isdigit(c); c = getchar()) {
        x = x * 10 + c - '0';
    }
    if (f) {
        x = -x;
    }
}

#define LL long long

LL n, x, y, ans;
int q;

signed main() {
	read(n);
	if(n & 1) {
		read(q);
		while(q--) {
			ans = 0;
			read(x), read(y);
			if(x & 1) {
				if(y & 1) {
					ans += (x >> 1) * (n >> 1) + (x >> 1) * ((n >> 1) + 1);
					ans += ((y + 1) >> 1);
				}
				else {
					ans += (n * n / 2) + 1;
					ans += (x >> 1) * (n >> 1) + (x >> 1) * ((n >> 1) + 1);
					ans += (y >> 1);
				}
			}
			else {
				if(y & 1) {
					ans += (n * n / 2) + 1;
					ans += (x >> 1) * (n >> 1) + ((x >> 1) - 1) * ((n >> 1) + 1);
					ans += ((y + 1) >> 1);
				}
				else {
					ans += (x >> 1) * ((n >> 1) + 1) + ((x >> 1) - 1) * (n >> 1);
					ans += (y >> 1);
				}
			}
			printf("%I64d\n", ans);
		}
	}
	else {
		read(q);
		while(q--) {
			ans = 0;
			read(x), read(y);
			if(x & 1) {
				if(y & 1) {
					ans += (x - 1) * (n >> 1);
					ans += ((y + 1) >> 1); 
				}
				else {
					ans += (n * n / 2);
					ans += (x - 1) * (n >> 1);
					ans += (y >> 1);
				}
			}
			else {
				if(y & 1) {
					ans += (n * n / 2);
					ans += (x - 1) * (n >> 1);
					ans += ((y + 1) >> 1);
				}
				else {
					ans += (x - 1) * (n >> 1);
					ans += (y >> 1);
				}
			}
			printf("%I64d\n", ans);
		}
	}
	return 0;
}

C,

大意是从一些长度的木棍中选出四根组成一个矩形，要求矩形的周长平方除以面积的商最小，求长宽x和y

将这个式子写成(2x + 2y) ^ 2 / xy，转化一下，由均值不等式得当x==y时该式最小

然后，假设a < b < c, 设上式为一个函数f(x, y)，则f(a, b) < f(a, c)，f(b, c) < f(a, c)

所以将所有有至少两根的木棍长度排序，计算相领的两个长度的f(x, y)值，取最小

如果有4个一样长度的直接输出

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 1000100;
const int INF = 0x3f3f3f3f;

template <typename T> inline void read(T &x) {
    int c = getchar();
    bool f = false;
    for (x = 0; !isdigit(c); c = getchar()) {
        if (c == '-') {
            f = true;
        }
    }
    for (; isdigit(c); c = getchar()) {
        x = x * 10 + c - '0';
    }
    if (f) {
        x = -x;
    }
}

#define LL long long

int T, n, l1, l2;
int a[MAXN], sum[MAXN / 30];
vector<int> v;

signed main() {
	read(T);
	while(T--) {
		read(n);
		bool fg = 0;
		memset(sum, 0, sizeof(sum));
		v.clear();
		for(int i = 1; i <= n; i++) {
			read(a[i]);
			sum[a[i]]++;
			if(sum[a[i]] == 4 && !fg) {
				printf("%d %d %d %d\n", a[i], a[i], a[i], a[i]);
				fg = 1;
			}
			if(sum[a[i]] == 2) {
				v.push_back(a[i]);
			}
		}
		if(fg) continue;
		sort(v.begin(), v.end());
		double dlt = 1e50;
		for(int i = 0; i < (int) v.size() - 1; i++) {
			double x = (double) v[i], y = (double) v[i + 1];
			double tmp = 8.000 + 4.000 * (x / y + y / x);
			if(tmp < dlt) {
				l1 = v[i], l2 = v[i + 1];
				dlt = tmp;
			}
		}
		printf("%d %d %d %d\n", l1, l2, l2, l1);
	}
	return 0;
}

D

一道tarjan的题，大概相当于求一个有向图中所有的出度为0的SCC，没什么好说的，看出来了之后做法就很粗暴了，具体可参看popular cow

#include <bits/stdc++.h>
using namespace std;

const int MAXM = 10010;
const int INF = 0x3f3f3f3f;

template <typename T> inline void read(T &x) {
    int ch = getchar();
    bool f = false;
    for (x = 0; !isdigit(ch); ch = getchar()) {
        if (ch == '-') {
            f = true;
        }
    }
    for (; isdigit(ch); ch = getchar()) {
        x = x * 10 + ch - '0';
    }
    if (f) {
        x = -x;
    }
}

#define LL long long

const int MAXN = 200100;
const int MAXE = 600400;

using std::min;

struct Edge {
    int to, nxt;
    Edge() {}
    Edge(int _to, int _nxt) : to(_to), nxt(_nxt) {}
}E[MAXE];

int h[MAXN], ins[MAXN], dfn[MAXN], Clock_t, Stack[MAXN], SCC;
int top, belong[MAXN], in[MAXN], sz[MAXN], cnt, low[MAXN];
int vis[MAXN], n, m;

inline void add_edge(int u, int v) {
    E[++cnt] = Edge(v, h[u]), h[u] = cnt;
    //E[++cnt] = Edge(u, h[v]), h[v] = cnt;
}

int a[MAXN], c[MAXN];

void Tarjan(int x) {
    ins[x] = 1;
    low[x] = dfn[x] = ++Clock_t;
    Stack[++top] = x;
    for(int i = h[x]; ~i; i = E[i].nxt) {
        int v = E[i].to;
        if(!dfn[v]) {
            Tarjan(v);
            low[x] = min(low[x], low[v]);
        }
        else if(ins[v]) low[x] = min(low[x], low[v]);
    }
    if(low[x] == dfn[x]) {
        SCC++;int v;
        do {
            v = Stack[top--];
            ins[v] = 0;
            belong[v] = SCC;
            sz[SCC] = min(sz[SCC], c[v]);
        }while(v != x);
    }
    return ;
}

int out[MAXN];

signed main() {
	read(n); memset(h, -1, sizeof(h)); memset(sz, 0x3f, sizeof(sz));
	for(int i = 1; i <= n; i++) read(c[i]);
	for(int i = 1; i <= n; i++) read(a[i]), add_edge(i, a[i]);
	for(int i = 1; i <= n; i++) {
		if(!dfn[i])
			Tarjan(i);
	}
	int ans = 0;
	for(int i = 1; i <= n; i++) {
		for(int x = h[i]; ~x; x = E[x].nxt) {
			int v = E[x].to;
			if(belong[i] != belong[v]) out[belong[i]]++; 
		}
	}
	for(int i = 1; i <= SCC; i++) {
		if(!out[i]) ans += sz[i];
	}
	printf("%d\n", ans);
	return 0;
}