Educational Codeforces Round 103 (Rated for Div. 2)A~E解题报告

Educational Codeforces Round 103 (Rated for Div. 2)

A. K-divisible Sum

原题信息

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解题思路

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AC代码

#include <bits/stdc++.h>
using namespace std;typedef long long LL;
const int N = 100010;int main()
{int t;  cin >> t;while (t -- ){static LL n, k;scanf("%lld%lld", &n, &k);k = (n / k + (n % k != 0)) * k;// printf("N=%lld, k=%lld\n", n, k);cout << (k / n + (k % n != 0)) << endl;}return 0;
}

B. Inflation

原题信息

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解题思路

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AC代码

#include <bits/stdc++.h>
using namespace std;typedef long long LL;
const int N = 100010;
int n, k;
LL p[N], b[N];int main()
{int t;  cin >> t;while (t -- ){scanf("%d%d", &n, &k);for (int i = 0; i < n; i ++ )scanf("%lld", &p[i]);b[0] = p[0];for (int i = 1; i < n; i ++ )b[i] = b[i - 1] + p[i];LL res = 0;for (int i = 1; i < n; i ++ ){res = max(res, 100LL * p[i] / k - b[i - 1] + (100LL * p[i] % k != 0));}cout << res << endl;}return 0;
}

C. Longest Simple Cycle

原题信息

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解题思路

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AC代码

#include <bits/stdc++.h>
using namespace std;typedef long long LL;
const int N = 100010;
const LL INF = 0x3f3f3f3f3f3f3f3f;
int n, k;
LL res;
LL a[N], b[N], c[N];int main()
{int t;  cin >> t;while (t -- ){res = 0;// inputscanf("%d", &n);for (int i = 1; i <= n; i ++ )  scanf("%lld", &c[i]);for (int i = 1; i <= n; i ++ )  scanf("%lld", &a[i]);for (int i = 1; i <= n; i ++ )  scanf("%lld", &b[i]);for (int i = 1; i <= n; i ++ )if (a[i] > b[i])swap(a[i], b[i]);LL tmp, pre = -INF;for (int i = 2; i <= n; i ++ ){if (a[i] == b[i])tmp = c[i] + 1;elsetmp = max(c[i] + b[i] - a[i] + 1, c[i] + pre - (b[i] - a[i] - 1));res = max(res, tmp);pre = tmp;}cout << res << endl;}return 0;
}

D. Journey

原题信息

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解题思路

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AC代码

#include <bits/stdc++.h>
using namespace std;const int N = 301010;
int f_l[N], f_r[N];
int ans[N];
char s[N];
int n;int main()
{int t;  cin >> t;while (t -- ){scanf("%d", &n);scanf("%s", s + 1);// leftf_l[1] = 1;for (int i = 2; i <= n; i ++ )if (s[i] != s[i - 1])f_l[i] = f_l[i - 1] + 1;elsef_l[i] = 1;// rightf_r[n] = 1;for (int i = n - 1; i >= 1; i -- )if (s[i] != s[i + 1])f_r[i] = f_r[i + 1] + 1;elsef_r[i] = 1;// 初始化并计算结果memset(ans, 0, sizeof ans);for (int i = 1; i <= n; i ++ )if (s[i] == 'R')ans[i - 1] += f_r[i];elseans[i] += f_l[i];cout << ans[0] + 1;for (int i = 1; i <= n; i ++ )printf(" %d", ans[i] + 1);puts("\n");}return 0;
}

E. Pattern Matching

原题信息

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解题思路

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倘若题目的要求，变一下，变成 rearrange 后的 pattern 在 mt[j] 的位置上是与 str[j] 匹配的，那么本题就变成了一个 二分图匈牙利算法的问题
显示对 m 个 str进行排序，按照他们的 mt进行排序，最后看每个mt需要满足的匹配要求，连向可以连接的 pattern，这样二分图就构建完成，直接跑一边匈牙利算法即可。
这是我之前一直读错的题意 ，绝了。

AC代码

#include <bits/stdc++.h>
using namespace std;const int BASE = 27, M = 27 * 27 * 27 * 27 + 10;
const int N = 100010, K = 7;
int _hash[M], char_num[258];   // _hash 需要开辟的数组
int n, m, k;
char _str[N][K];        // n 个 pattern
char pattern_match[K];  // dfs查询 m 个 str可以匹配的 pattern 可能
char str[5]; int mt;    // m 个询问
vector<int> idx_ret;    // 记录返回的编号
int h[N], e[M * 5], ne[M * 5], idx;
bool flag = true;       // 全局变量，记录是否可以
int ind[N];
int res[N];             // 存储最终的排序结果// 计算字符串hash的数值
int Cal(char *s)
{int ret = 0;for (int i = 0, base = 1; i < k; i ++, s ++){ret += base * char_num[*s];base = base * BASE;}return ret;
}// 第一个是 str， 第二个是 pattern，可以带有_
bool Match(char *s, char *pa)
{for (int i = 0; i < k; i ++ ){if (*s == *pa || *pa == '_' || *s == '_'){s ++;   pa ++;continue;}return false;}return true;
}void GetIdx(char *s, int cnt)
{if (cnt == k){if (Match(str, pattern_match)){if (_hash[Cal(pattern_match)] != -1){idx_ret.push_back(_hash[Cal(pattern_match)]);}}return;}else{pattern_match[cnt] = *s;GetIdx(s + 1, cnt + 1);pattern_match[cnt] = '_';GetIdx(s + 1, cnt + 1);}
}void add(int x, int y)
{e[idx] = y, ne[idx] = h[x], h[x] = idx ++;
}void topsort()
{int cur_idx = 1;queue<int> que;for (int i = 1; i <= n; i ++ ){if (ind[i] == 0)que.push(i);}int u, v;/// printf("TOPSORT:\n");while (que.size()){u = que.front();    que.pop();res[cur_idx ++] = u;/// printf("%d, ", u);for (int i = h[u]; ~i; i = ne[i]){v = e[i];ind[v] --;if (ind[v] == 0)que.push(v);}}/// cout << endl;/// printf("n=%d, curidx=%d\n", n, cur_idx);flag = (cur_idx == n + 1);/// cout << flag << endl;
}int main()
{// _hash initialmemset(_hash, -1, sizeof _hash);for (int i = 'a'; i <= 'z'; i ++ )char_num[i] = i - 'a';char_num['_'] = 26;// edge list initialmemset(h, -1, sizeof h);idx = 1;// input and build string hashcin >> n >> m >> k;for (int i = 1; i <= n; i ++ ){scanf("%s", _str[i]);_hash[Cal(_str[i])] = i;}
/*cout << 27 * 27 * 27 * 27 - 1 << endl;cout << Cal("abc_") << endl;cout << Cal("____") << endl;cout << Cal("aaaa") << endl;return 0;
*/// ind initialmemset(ind, 0, sizeof ind);for (int i = 1; flag && i <= m; i ++ ){scanf("%s%d", str, &mt);if (!Match(str, _str[mt])){puts("NO");return 0;}else{idx_ret.clear();GetIdx(str, 0);/// printf("\t*J:%d\n", i);for (int j = 0; flag && j < idx_ret.size(); j ++){if (mt != idx_ret[j]){add(mt, idx_ret[j]);ind[idx_ret[j]] ++;}/// cout << idx_ret[j] << ", ";}/// puts("");}}//if (!flag){/// printf("BEGIN\n");puts("NO");}else{topsort();if (!flag){puts("NO");}else{puts("YES");cout << res[1];for (int i = 2; i <= n; i ++ )printf(" %d", res[i]);puts("");}}return 0;
}