太久没做题就会变得很菜。
题目链接：https://codeforces.com/contest/1281/
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 }
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<<'\n';
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<<'\n';
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 }
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 << '\n';
46 }
47 return 0;
48 }
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\n", ans);
70 }
71 return 0;
72 }
来源：https://www.cnblogs.com/JHSeng/p/12045571.html