Codeforces Round #607 (Div. 2)

Posted 2021-03-22 jhseng

太久没做题就会变得很菜。

题目链接：https://codeforces.com/contest/1281/

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A：

白给。

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i<b;++i)
11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 int t;
21 const int maxn=1010;
22 char s[maxn];
23 
24 int main() {
25     scanf("%d",&t);
26     while(t--){
27         scanf("%s",s+1);
28         int len=strlen(s+1);
29         if (s[len]==‘o‘) puts("FILIPINO");
30         else if (s[len]==‘u‘) puts("JAPANESE");
31         else puts("KOREAN");
32     }
33     return 0;
34 }

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B：

显然对于每一位，只能跟后面的字符交换。如果只是O(n)扫，遇到第一个s[i]>c[i]的情况才考虑替换的话是错的。应该先把s的副本sort一遍，这样得到的结果是s每个位置交换能得到的最小的字符。当s[i]>s_copy[i]，即当前字符有优化空间时就立即考虑替换。

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i<b;++i)
11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (minpos<<1)
15 #define rson (minpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 int t;
21 string a,b;
22 
23 int main() {
24     cin>>t;
25     while (t--){
26         cin>>a>>b;
27         if (a<b){
28             cout<<a<<‘
‘;
29             continue;
30         }
31         string s=a;
32         sort(s.begin(),s.end());
33         int flag=0;
34         for (int i=0;i<(int)a.size();i++){
35             // current char is not at the best position
36             if (a[i]>s[i]){
37                 // it can swap with the back char only
38                 for (int j=i+1;j<(int)a.size();j++){
39                     swap(a[i],a[j]);
40                     if (a<b){
41                         cout<<a<<‘
‘;
42                         flag=1;
43                         i=(int)a.size();
44                         break;
45                     }
46                     swap(a[i],a[j]);
47                 }
48             }
49         }
50         if (!flag) puts("---");
51     }
52     return 0;
53 }

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C：

看似很数学，其实直接模拟就过了。注意当前字符串长度大于等于x时则停止延长字符串即可。

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i<b;++i)
11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 const int mod = 1e9 + 7;
21 int t;
22 
23 int main() {
24     ios::sync_with_stdio(false);
25     cin.tie(0);
26 
27     cin >> t;
28     while (t--) {
29         int x; string s; cin >> x >> s;
30         int currLen = (int)s.size();
31         for (int i = 0; i < x; i++) {
32             int curr = s[i] - ‘0‘;
33             // length need to grow
34             if (currLen < x) {
35                 for (int j = 1; j < curr; j++)
36                     if ((int)s.size() < x) {
37                         for (int l = i + 1; l < currLen; l++)
38                             if ((int)s.size() < x) s += s[l];
39                     }
40             }
41             // calculate current length of string
42             int d = (currLen - i - 1 + mod) % mod;
43             currLen = ((i + 1) + (ll)d * curr) % mod;
44         }
45         cout << currLen << ‘
‘;
46     }
47     return 0;
48 }

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D：

分类讨论题。很显然答案只能在[0,4]之间取值。枚举每一行每一列，对于每一行，检查其是否为全A、A在边缘和A在内部的情况，维护答案。列同理。

 1 /* basic header */
 2 #include <bits/stdc++.h>
 3 /* define */
 4 #define ll long long
 5 #define dou double
 6 #define pb emplace_back
 7 #define mp make_pair
 8 #define sot(a,b) sort(a+1,a+1+b)
 9 #define rep1(i,a,b) for(int i=a;i<=b;++i)
10 #define rep0(i,a,b) for(int i=a;i<b;++i)
11 #define eps 1e-8
12 #define int_inf 0x3f3f3f3f
13 #define ll_inf 0x7f7f7f7f7f7f7f7f
14 #define lson (curpos<<1)
15 #define rson (curpos<<1|1)
16 /* namespace */
17 using namespace std;
18 /* header end */
19 
20 const int maxn = 100;
21 int t, n, m;
22 char a[maxn][maxn];
23 
24 int main() {
25     scanf("%d", &t);
26     while (t--) {
27         int foundA = 0, foundP = 0, ans = 4;
28         scanf("%d%d", &n, &m);
29         for (int i = 1; i <= n; i++) {
30             scanf("%s", a[i] + 1);
31             for (int j = 1; j <= m; j++)
32                 if (a[i][j] == ‘A‘) foundA = 1;
33                 else foundP = 1;
34         }
35         if (!foundA) {
36             puts("MORTAL");
37             continue;
38         }
39         if (!foundP) {
40             puts("0");
41             continue;
42         }
43         for (int i = 1; i <= n; i++) {
44             int maxx = -1, minn = 1000;
45             for (int j = 1; j <= m; j++) {
46                 maxx = max(maxx, (int)a[i][j]), minn = min(minn, (int)a[i][j]);
47                 if (a[i][j] == ‘A‘) {
48                     int t = 4;
49                     if (i == 1 || i == n) t--;
50                     if (j == 1 || j == m) t--;
51                     ans = min(ans, t);
52                 }
53             }
54             if (maxx == minn && maxx == ‘A‘) {
55                 if (i == 1 || i == n) ans = min(ans, 1);
56                 else ans = min(ans, 2);
57             }
58         }
59         for (int j = 1; j <= m; j++) {
60             int maxx = -1, minn = 1000;
61             for (int i = 1; i <= n; i++) {
62                 maxx = max(maxx, (int)a[i][j]), minn = min(minn, (int)a[i][j]);
63             }
64             if (maxx == minn && maxx == ‘A‘) {
65                 if (j == 1 || j == m) ans = min(ans, 1);
66                 else ans = min(ans, 2);
67             }
68         }
69         printf("%d
", ans);
70     }
71     return 0;
72 }

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