字符串类型

代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int n;
 5 int cnt[500];
 6 string s;
 7 int mian() {
 8     cin>>s;
 9     n=s.size();
10     s=' '+s;
11 
12     for(int i=1; i<=n; i++) {
13         cnt[s[i]]++;
14     }
15 
16     int ok=1;
17     for(int i='a'; i<='z'; i++) {
18         if(cnt[i]%2==1) ok=0;
19     }
20     if(ok) cout<<"Yes" ;
21     else cout<<"No";
22 }

View Code

2.B - Varied (atcoder.jp)

 1 #include <iostream>
 2 #include <cstring>
 3 using namespace std;
 4 
 5 
 6 
 7 int main()
 8 {
 9     string s;
10     cin >> s;
11     int  i, j;
12     
13     for (i = 0; i < s.length(); i++)
14     {
15         for (j = i+1; j < s.length(); j++)
16         {
17             if (s[j] == s[i])
18             {
19                 printf("no");
20                 return 0;
21             }
22         }
23     }
24     
25     printf ("yes");
26     return 0;
27 }

View Code

 1 #include <stdio.h>
 2 #include <string.h>
 3 int main(){
 4     char a[26];
 5     int b[26]={0};
 6     scanf("%s",&a);
 7     int n;
 8     n=strlen(a);
 9     for(int i=0;i<n;i++){
10         b[a[i]-97]++;
11     }
12     int flag=1;
13     for(int i=0;i<n;i++){
14         if(b[i]>1){
15             flag=0;
16         }
17     }
18     if(flag==0){
19         printf("no");
20     }
21     else{
22         printf("yes");
23     }
24     return 0;
25 }

View Code

byd第三个代码就是过不了

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 string a;
 5 int s,cnt[1000];
 6 int main()
 7 {
 8    cin >> a;
 9    s = a.size();
10    
11    for(int i = 0; i < s ; i ++)
12    {
13        cnt[a[i]] ++;
14    }
15    
16    int ok = 1;
17    for(int i = 0; i < n; i ++)
18    {
19        if(cnt[i] > 1) ok = 0;
20    }
21    
22    if(ok) cout << "Yes";
23    else cout << "No";
24    return 0 ;
25 }

View Code

3.Increment Decrement - AtCoder abc052_b - Virtual Judge (vjudge.net)

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 int n,cnt,ma;
 5 char a[1000];
 6 
 7 
 8 int main()
 9 {
10     cin >> n;
11     
12     for (int i = 1; i <= n; i ++ )
13     {
14         cin >> a[i];
15         if(a[i] == 'D') cnt --;
16         else cnt ++;
17         if(cnt > ma) ma = cnt;
18     }
19     cout << ma;
20     return 0;
21 }

View Code

4.字符串的增删Unhappy Hacking (ABC Edit) - AtCoder abc043_b - Virtual Judge (vjudge.net)

for(char x: s) 用pop_back()和push_back()

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main(){
 4     ios::sync_with_stdio(false);
 5     cin.tie(0);
 6 
 7     string s,tmp="";
 8     cin>>s;
 9     
10     for(char x:s){
11         if(x=='B'&&!tmp.empty()){
12             tmp.pop_back();
13         }
14         else if(x!='B'){
15             tmp.push_back(x);
16         }
17     }
18     cout<<tmp<<"\n";
19     return 0;
20 }

View Code

5.Two Anagrams - AtCoder abc082_b - Virtual Judge (vjudge.net)

字符串的排序和逆转

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main() {
 4     string s,t;
 5     
 6     cin>>s;
 7     cin>>t;
 8     
 9     sort(s.begin(),s.end());
10     sort(t.begin(),t.end());
11     reverse(t.begin(),t.end());
12     
13     if(s<t)
14         cout<<"Yes";
15     else
16         cout<<"No";
17     return 0;
18 }

View Code

6.Grid Compression - AtCoder abc107_b - Virtual Judge (vjudge.net)

字符串组的整排，整列的增删，如果一排都相等，则重新定义一个数组去进行判断

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 int n,m;
 5 char s[101][101];
 6 int l[101],r[101];
 7 
 8 int main()
 9 {
10     cin >> n >> m;
11     for (int i = 1; i <= n; i ++ )
12        for (int j = 1; j <= m; j ++ ) cin >>s[i][j];
13 
14     for (int i = 1; i <= n; i++){
15         int t = 0;
16         for (int j = 1; j <= m; j++){
17             if(s[i][j] == '.'){
18                 t++;
19             }
20         }
21         if(t == m) l[i] = 1;
22     }
23     for (int i = 1; i <= m; i++){
24         int t = 0;
25         for (int j = 1; j <= n; j++){
26             if(s[j][i] == '.'){
27                 t++;
28             }
29         }
30         if(t == n) r[i] = 1;
31     }
32     for (int i = 1; i <= n; i++){
33         int flag = 0;
34         for (int j = 1; j <= m; j++){
35             if(l[i] || r[j]) continue;
36             flag = 1;
37             printf("%c", s[i][j]);
38         }
39         if(flag) printf("\n");
40     }
41  
42     return 0;
43 }

View Code

7.字符串组的增删Snuke's Coloring 2 (ABC Edit) - AtCoder abc047_b - Virtual Judge (vjudge.net)

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 
 5 int main(){
 6     ll n,m,q;
 7     cin >> m >> n >> q;
 8     
 9     ll s[n+5][m+5];
10     for(ll i = 1;i <= n;i++){
11         for(ll j = 1;j <= m;j++){
12             s[i][j] = 1;
13         }
14     }
15     while(q--){
16         ll x,y,a;
17         cin>>x>>y>>a;
18         if(a == 1){
19             for(ll i=1;i<=n;i++){
20                 for(ll j=1;j<=x;j++) s[i][j]=0;
21             }
22         }
23         if(a==2){
24             for(ll i=1;i<=n;i++){
25                 for(ll j=x+1;j<=m;j++) s[i][j]=0;
26             }
27         }
28         if(a==3){
29             for(ll i=n-y+1;i<=n;i++){
30                 for(ll j=1;j<=m;j++) s[i][j]=0;
31             }
32         }
33         if(a==4){
34             for(ll i=1;i<=n-y;i++){
35                 for(ll j=1;j<=m;j++) s[i][j]=0;
36             }
37         }
38     }
39     ll res=0;
40     for(ll i=1;i<=n;i++){
41         for(ll j=1;j<=m;j++){
42             if(s[i][j]) res++;
43         }
44     }
45     cout<<res;
46 }

View Code

8.表示第一次出现和最后一次出现的新颖的方法，以及找规律

找到第一个A出现的位置和最后一次B出现的位置

Problem - B - Codeforces

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int N = 2e5 + 10;
 5 int n,t;
 6 char s[N];
 7 
 8 int main()
 9 {
10     cin >> t;
11     while(t -- )
12     {
13         scanf("%d%s", &n, s + 1);
14         
15         int ans = 0,x = 0,y = 0;
16         for (int i = 1; i <= n; i ++ )
17         {
18             if(s[i] == 'B') y = i;
19             else {
20                 if(x == 0) x = i;
21             }
22         }
23         if(x!=0&&y!=0){
24             if(y >= x) ans = y - x;
25         }
26         cout << ans << endl;
27     }
28 }

Code

9.又又又是字符串的排序和逆转 Problem - C - Codeforces

这个题要求找到最长的非上升字典序进行循环排序，找到最少操作次数，我们可以把按照要求的字符下标丢到vetor里面然后swap转换，最后再判断一下是否有序

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define endl '\n'
 4 #define int long long
 5 #define cy cout << "YES" << endl
 6 #define cn cout << "NO" << endl
 7 #define c1 cout << -1 << endl
 8 int _,n,m,cnt;
 9 const int N = 4e3 + 10,inf = 1e9;
10 const int mod = 1e9 + 7;
11 
12 signed main()
13 {
14     cin >> _;
15     while(_ -- ){
16         int n;
17         cin >> n;
18         string s;
19         cin >> s;
20         vector<int> a;
21         for (int i = n - 1; i >= 0; i -- )
22            if(a.empty() || s[i] >= s[a.back()]) //如果后一位字典序小于等于前一位
23            a.push_back(i); //存入最大子序列下标
24         
25         reverse(a.begin(),a.end());
26         int ans = 0;
27         int j = 0;
28         while(j < a.size() && s[a[0]] == s[a[j]]) j += 1;
29         ans = a.size() - j;
30         
31         for (int i = 0; i < a.size() - 1 - i; i ++ ){
32             swap(s[a[i]],s[a[a.size() - 1 - i]]); //循环子序列
33         }
34         
35         if(!is_sorted(s.begin(),s.end())) ans = -1;//判断循环后的序列是否有序
36         cout << ans << endl;
37     }
38     return 0;
39 }

Code

posted @ 2023-11-04 16:21 rw156 阅读(11) 评论(0) 编辑收藏举报

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