Educational Codeforces Round 45 (Rated for Div. 2)

A Commentary Boxes

算出$N \pmod M$

然后分别讨论是加还是减

#include<cstdio>
#include<algorithm>
#include<map>
#include<vector>
#define int long long
using namespace std;
inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;
}
int N, M, A, B;
main() {N = read(); M = read(); A = read(), B = read();int res = N % M;printf("%lld", min((M - res) * A, res * B));
}

B Micro-World

排序之后用一个栈维护当前的病毒

然后不断吞就好了

#include<cstdio>
#include<algorithm>
#include<map>
#include<vector>
#define int long long
using namespace std;
const int MAXN = 1e6 + 10;
inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;
}
int N, K;
int a[MAXN], s[MAXN], top = 0;
main() {N = read(); K = read();for(int i = 1; i <= N; i++) a[i] = read();sort(a + 1, a + N + 1);int ans = 0;s[++top] = a[1];for(int i = 2; i <= N; i++) {while(top > 0 && (a[i] > s[top]) && (a[i] <= s[top] + K)) ans++, top--;s[++top] = a[i];}printf("%d", N - ans);
}

C Bracket Sequences Concatenation Problem

类似于$)($的肯定是不管与谁都无法匹配

$(($这样的只能匹配$))$，反过来同理

$()$这样合法的于合法的都能匹配

然后记录一下乘法原理统计答案

#include<cstdio>
#include<algorithm>
#include<map>
#include<vector>
#include<cstring>
#define LL long long
using namespace std;
const int MAXN = 3e5 + 10;
int N, M, A, B;
int s[MAXN];
LL top = 0, L[MAXN], R[MAXN], can, limit = 0;
char ss[MAXN];
main() {//freopen("a.in", "r", stdin);//freopen("c.out", "w", stdout);int N;scanf("%d", &N);for(int i = 1; i <= N; i++) {scanf("%s", ss + 1);top = 0;int Len = strlen(ss + 1);bool flag = 0;for(int j = 1; j <= Len; j++)    {if(ss[j] == '(') s[++top] = 1;if(ss[j] == ')') {if(top > 0 && s[top] == 1) top--;else s[++top] = 2;}}for(int j = 2; j <= top; j++)if(s[j] != s[j - 1]){flag = 1; break;}if(flag == 1) continue;limit = max(limit, top);if(top == 0) {L[0]++; R[0]++; continue;}if(s[top] == 2) R[top]++;if(s[top] == 1) L[top]++;}LL ans = 0;for(int i = 0; i <= limit; i++)ans += 1ll * L[i] * R[i];printf("%lld\n", ans);}

D Graph And Its Complement

第一次遇到构造题

首先当$a,b$同时不唯一的时候一定是无解的，

证明：假设原图的联通块数不为$1$，那么原图中不同联通块之间的点在反图中一定有边，

这样原图中的不同联通块之间的点可以通过新边到达其他联通块，

然后再通过其他联通块的新边回到原来的联通块

这样我们假设$a=1$（若不满足可以交换）

设$b=x$

首先我们把原图中的边全都连上，然后在原图中任意$N-x$对点的边断开就好

注意要特判$N=2$和$N=3$的情况

#include<cstdio>
#include<algorithm>
#include<map>
#include<vector>
#define LL long long
using namespace std;
const LL MAXN = 1001, INF = 1e18 + 10;
inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;
}
char ans[MAXN][MAXN];
main() {int N = read(), A = read(), B = read();if(A != 1 && B != 1) { printf("NO"); return 0;}char a = '1', b = '0'; if(A != 1) swap(a, b), swap(A, B);if((N == 2 || N == 3) && B == 1) { printf("NO"); return 0;}printf("YES\n");for(int i = 1; i <= N; i++)for(int j = 1; j <= N; j++)if(i != j) ans[i][j] = a;else ans[i][j] = '0';for(int i = 1; i <= (N - B); i++) ans[i][i + 1] = b, ans[i + 1][i] = b;for(int i = 1; i <= N; i++, puts(""))for(int j = 1; j <= N; j++)putchar(ans[i][j]);
}

E Post Lamps

预处理出障碍，

直接暴力枚举选哪个

时间复杂度为调和级数$O(klogn)$

#include<cstdio>
#include<algorithm>
#include<map>
#include<vector>
#define LL long long
using namespace std;
const LL MAXN = 1e6 + 10, INF = 1e18 + 10;
inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;
}
int N, M, K;
int block[MAXN], sum[MAXN], a[MAXN], cnt[MAXN];
main() {N = read(); M = read(); K = read();for(int i = 1; i <= M; i++) block[read()] = 1;if(block[0]) {printf("-1"); return 0;}for(int i = 1; i <= K; i++) a[i] = read();int mx = 0;for(int i = 1, j; i <= 1e6; i = j + 1) {j = i; int num = 0;while(block[j]) mx = max(cnt[j] = ++num, mx), j++;}LL out = INF;for(int i = mx + 1; i <= K; i++) {LL now = 0, spend = 0;while(now < N) {if(block[now]) now = now - cnt[now];else now = now + i, spend = spend + a[i];}out = min(out, spend);}printf("%lld", out == INF ? -1 : out);
}

总结

第一次打cf，确实有很多不适应的地方，第一题上来把$n$和$m$看反了，然后特判的时候写的是$M % N$，直接wa到飞

T2秒的比较快

T3也秒的比较快，不过写代码的时候把$)($判成了$()$，又wa成傻逼。

T4没看，不过zbq秒了(不过他考场上没判$n=3$..)，然后赛后做了做

T5最后几分钟看了看，当时感觉比较可做，但是思路一直纠结在如何处理障碍上，我一直以为障碍的范围是$10^9$然后纠结要不要开map

怎么说呢，感觉自己最近真的太浮躁了，很多时候连范围都没看清楚就开始做。

希望自己往后打cf的时候能沉下心来一句一句的读题目吧。

转载于:https://www.cnblogs.com/zwfymqz/p/9165354.html