Codeforces Round #327 (Div. 2)-白红宇

Codeforces Round #327 (Div. 2)

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发布时间：2019-06-12

本文共 3205 字，大约阅读时间需要 10 分钟。

水

题目都没看清就写了，1e-4精度WA了一次。。。

/************************************************* Author        :Running_Time* Created Time  :2015/10/25 16:27:20* File Name     :A.cpp ************************************************/#include 
     
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                      using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1typedef long long ll;const int N = 1e5 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;int main(void) { int L, p, q; scanf ("%d%d%d", &L, &p, &q); double ans = L * (p * 1.0) / (p + q); printf ("%.5f\n", ans); return 0;}

构造

题意：要求字符串的所有C1字符变成C2，C2变成C1，输出最后的结果

分析：想了一会，试了并查集，未果，YY，未果。最后想了一个很奇怪的方法，就是每次记录C1的最原始的字符rt[C1]，它将转换为C2，即to[rt[C1]] = C2

/************************************************* Author        :Running_Time* Created Time  :2015/10/25 16:27:20* File Name     :B.cpp ************************************************/#include 
     
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                      using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1typedef long long ll;const int N = 2e6 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;char s[N];int to[30], rt[30];int main(void) { int n, m; scanf ("%d%d", &n, &m); getchar (); scanf ("%s", &s); getchar (); char c1, c2; memset (to, -1, sizeof (to)); for (int i=0; i<26; ++i) rt[i] = i; for (int i=1; i<=m; ++i) { scanf ("%c %c", &c1, &c2); getchar (); int tmp = rt[c2-'a']; to[rt[c1-'a']] = c2 - 'a'; to[rt[c2-'a']] = c1 - 'a'; rt[c2-'a'] = rt[c1-'a']; rt[c1-'a'] = tmp; } for (int i=0; i

找规律

题意：由01构成的序列，每一次a[i] = (a[i-1], a[i], a[i+1])的第二大，问多少次序列会稳定

分析：列出(a[i-1], a[i], a[i+1])的所有组合，发现只有010和101是不稳定的，所以找出这样的连续的最长的串，操作次数就是max_len / 2

/************************************************* Author        :Running_Time* Created Time  :2015/10/25 16:27:20* File Name     :C.cpp ************************************************/#include 
     
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                      using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1typedef long long ll;const int N = 5e5 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;int a[N];int main(void) { int n; scanf ("%d", &n); for (int i=1; i<=n; ++i) { scanf ("%d", &a[i]); } int ans = 0; for (int i=2; i
                      
                       > 1); int p = i, q = j - 1; while (p <= q) { a[p++] = a[i-1]; a[q--] = a[j]; } i = j; } } printf ("%d\n", ans); for (int i=1; i<=n; ++i) { printf ("%d%c", a[i], i == n ? '\n' : ' '); } //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"; return 0;}

转载于:https://www.cnblogs.com/Running-Time/p/4911352.html

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