Codeforces Round #293 (Div. 2)
题意:两个字符串s,t,是否存在满足:s < r < t 的r字符串
字符转处理:字典序排序
很巧妙的方法,因为s < t,只要找比t字典序稍微小一点的和s比较就行了
具体方法和数字减1相类似,从"个位"减1,如果是0,从前面借1
!strcmp (t, s):如果t < s 或者 t > s (不可能)则输出,t == s 则输出NO
#include <cstdio> #include <iostream> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include <vector> #include <map> #include <set> using namespace std; const int MAXN = 1e6 + 10; const int INF = 0x3f3f3f3f; int main(void) { //freopen ("A.in", "r", stdin); char s[110], t[110]; while (~scanf ("%s %s", &s, &t)) { int cnt = strlen (s) - 1; while (t[cnt] == 'a' && cnt >= 0) { t[cnt] = 'z'; cnt--; } t[cnt]--; (!strcmp (t, s)) ? puts ("No such string") : printf ("%s\n", t); } return 0; }
字符串处理:字符查找
记录s,t各自的大小写字母的数量,然后累加完全匹配的cnty和不完全匹配的cntw
这道题题目我没读懂,cntw不完全匹配意思是:只是大小不相同
#include <cstdio> #include <iostream> #include <algorithm> #include <map> #include <cstring> #include <string> #include <set> using namespace std; const int MAXN = 2e5 + 10; const int INF = 0x3f3f3f3f; int main(void) { //freopen ("B.in", "r", stdin); char s[MAXN], t[MAXN]; int m1[30], m2[30], m3[30], m4[30]; while (cin >> s >> t) { memset (m1, 0, sizeof (m1)); memset (m2, 0, sizeof (m2)); memset (m3, 0, sizeof (m3)); memset (m4, 0, sizeof (m4)); for (int i=0; s[i]!='\0'; ++i) { if (s[i]>='a' && s[i]<='z') m1[s[i] - 'a']++; else m2[s[i]-'A']++; } for (int i=0; t[i]!='\0'; ++i) { if (t[i]>='a' && t[i]<='z') m3[t[i] - 'a']++; else m4[t[i]-'A']++; } int cnty = 0, cntw = 0; for (int i=0; i<26; ++i) { int d = min (m1[i], m3[i]); m1[i] -= d; m3[i] -= d; cnty += d; d = min (m2[i], m4[i]); m2[i] -= d; m4[i] -= d; cnty += d; } for (int i=0; i<26; ++i) { int d = min (m1[i], m4[i]); m1[i] -= d; m4[i] -= d; cntw += d; d = min (m2[i], m3[i]); m2[i] -= d; m3[i] -= d; cntw += d; } cout << cnty << " " << cntw << endl; } return 0; }
无算法
统计划屏的次数,如果在第一屏则不用,只要每次交换与前面数字的顺序就行了
注意:ans开long long
好吧,这道题是最水的,主要是题目很难读懂,可以从Note里猜出题目意思
#include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map> using namespace std; const int MAXN = 1e5 + 10; const int INF = 0x3f3f3f3f; int num[MAXN]; int pos[MAXN]; int main(void) { //freopen ("C.in", "r", stdin); int n, m, k; while (~scanf ("%d%d%d", &n, &m, &k)) { int x; for (int i=1; i<=n; ++i) { scanf ("%d", &x); pos[x] = i; num[i] = x; } long long ans = 0; for (int i=1; i<=m; ++i) { scanf ("%d", &x); int p = pos[x]; ans += (p / k); if (p % k) ans += 1; if (p == 1) continue; int y = num[p-1]; num[p-1] = x; num[p] = y; pos[y]++; pos[x]--; } printf ("%I64d\n", ans); } return 0; }
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