A - “atcoder”.substr()

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 100100100 points

Problem Statement

Print the LLL-th through RRR-th characters of the string atcoder.

Constraints

LLL and RRR are integers.
1≤L≤R≤71≤L≤R≤71≤L≤R≤7

Input

Input is given from Standard Input in the following format:

LLL RRR

Output

Print the answer.

题面翻译

给定左右边界LLL和RRR，求截取[L,R][L,R][L,R]后的atcoder字符串。

Sample Input 1

3 6

Sample Output 1

code

The 333-rd through 666-th characters of atcoder are code.

Sample Input 2

4 4

Sample Output 2

Sample Input 3

1 7

Sample Output 3

atcoder

题解部分

我们用一个string类型变量记录一下字符串atcoder，
再用一重循环输出区间内的字符即可。

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;int main () {string str = " atcoder";int l, r;scanf ("%d %d", &l, &r);for (int i = l; i <= r; i ++) {printf ("%c", str[i]);}return 0;
}

B - Nice Grid

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 200200200 points

Problem Statement

Print the color of the cell at the R-th row from the top and C-th column from the left in the following grid with 15 vertical rows and 15 horizontal columns.

Constraints

1≤R,C≤151≤R,C≤151≤R,C≤15
RRR and CCC are integers.

Input

Input is given from Standard Input in the following format:

RRR CCC

Output

In the grid above, if the color of the cell at the RRR-th row from the top and CCC-th column from the left is black, then print black; if the cell is white, then print white. Note that the judge is case-sensitive.

题面翻译

给出坐标(R,C)(R,C)(R,C)，判断上图第RRR行，第CCC列的颜色。

Sample Input 1

3 5

Sample Output 1

black

In the grid above, the cell at the 333-rd row from the top and 555-th column from the left is black. Thus, black should be printed.

Sample Input 2

4 5

Sample Output 2

white

In the grid above, the cell at the 444-th row from the top and 555-th column from the left is white. Thus, white should be printed.

题解部分

我们可以用用两重循环标记整个图的颜色，再输出对应字符串即可。

#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;bool fl[20][20];int main () {int a, b;scanf ("%d %d", &a, &b);for (int i = 1; i <= 7; i += 2) {for (int j = i; j <= 15 - i + 1; j ++) {//每一圈正方形的上、左边fl[i][j] = 1;fl[j][i] = 1;}for (int j = 15 - i + 1; j >= i; j --) {//每一圈正方形的下、右边fl[15 - i + 1][j] = 1;fl[j][15 - i + 1] = 1;}}printf ("%s", fl[a][b] == 1 ? "black" : "white");return 0;
}

C - Matrix Reducing

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 300300300 points

Problem Statement

You are given a matrix AAA with H1H_1H1 rows and W1W_1W1 columns, and a matrix BBB with H2H_2H2 rows and W2W_2W2 columns.

For all integer pairs (i,j)(i,j)(i,j) such that 1≤i≤H11≤i≤H_11≤i≤H1 and 1≤j≤W11≤j≤W_11≤j≤W1 , the element at the iii-th row and jjj-th column of matrix AAA is Ai,jA_{i,j}Ai,j .
For all integer pairs (i,j)(i,j)(i,j) such that 1≤i≤H21≤i≤H_21≤i≤H2 and 1≤j≤W21≤j≤W_21≤j≤W2, the element at the iii-th row and jjj-th column of matrix BBB is Bi,jB_{i,j}Bi,j.
You may perform the following operations on the matrix AAA any number of (possibly 000) times in any order:
Choose an arbitrary row of AAA and remove it.
Choose an arbitrary column of AAA and remove it.
Determine if it is possible to make the matrix AAA equal the matrix BBB.

Constraints

1≤H2≤H1≤101≤H_2≤H_1≤101≤H2≤H1≤10
1≤W2≤W1≤101≤W_2≤W_1≤101≤W2≤W1≤10
1≤Ai,j≤1091≤A_{i,j}≤10^91≤Ai,j≤109
1≤Bi,j≤1091≤B_{i,j}≤10^91≤Bi,j≤109
All values in input are integers.

Input

Input is given from Standard Input in the following format:

H1H_1H1 W1W_1W1
A1,1A_{1,1}A1,1 A1,2A_{1,2}A1,2 … A1,W1A_{1,W_1}A1,W1
A2,1A_{2,1}A2,1 A2,2A_{2,2}A2,2 … A2,W1A_{2,W_1}A2,W1
⋮
AH1,1A_{H_1,1}AH1,1 AH1,2A_{H_1,2}AH1,2 … AH1,W1A_{H_1,W_1}AH1,W1
H2H_2H2 W2W_2W2
B1,1B_{1,1}B1,1 B1,2B_{1,2}B1,2 … B1,W2B_{1,W_2}B1,W2
B2,1B_{2,1}B2,1 B2,2B_{2,2}B2,2 … B2,W2B_{2,W_2}B2,W2
⋮
BH2,1B_{H_2,1}BH2,1 BH2,2B_{H_2,2}BH2,2 … BH2,W2B_{H_2,W_2}BH2,W2

Output

Print Yes if it is possible to make the matrix AAA equal the matrix BBB; print No otherwise. Note that the judge is case-sensitive.

题面翻译

给出两个二维数组AAA、BBB，你可以删除若干次AAA的任意一行(列)，判断是否能将AAA转化为BBB。

Sample Input 1

4 5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
2 3
6 8 9
16 18 19

Sample Output 1

Yes

Removing the 222-nd column from the initial AAA results in:

1 3 4 5
6 8 9 10
11 13 14 15
16 18 19 20

Then, removing the 333-rd row from AAA results in:

1 3 4 5
6 8 9 10
16 18 19 20

Then, removing the 111-st row from AAA results in:

6 8 9 10
16 18 19 20

Then, removing the 444-th column from AAA results in:

6 8 9
16 18 19

Now the matrix equals the matrix BBB.
Thus, we can make the matrix AAA equal the matrix BBB by repeating the operations, so Yes should be printed.

Sample Input 2

Sample Output 2

No

Regardless of how we perform the operations, we cannot make the matrix AAA equal the matrix BBB, so No should be printed.

题解部分

注意数据范围，长和宽都≤10\le 10≤10，所以我们选择用两个dfs暴搜整个矩阵，如果两个矩阵完全一致，输出Yes，搜完了仍然没有输出，则输出No。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cstdlib>
using namespace std;const int MAXN = 10 + 5;
int a[MAXN][MAXN], b[MAXN][MAXN];
bool line[MAXN], col[MAXN];
int n1, m1, n2, m2;bool check () {//判断搜索后的矩阵A是否与B相等int I = 1, J;for (int i = 1; i <= n1; i ++) {if (col[i]) {J = 1;for (int j = 1; j <= m1; j ++) {if (line[j]) {if (a[i][j] != b[I][J]) {return 0;}J ++;}}I ++;}}return 1;
}void dfs_col (int step, int x) {//搜索每一列if (step > n2) {if (check ()) {printf ("Yes");exit (0);}return;}for (int i = x + 1; i <= n1 - (n2 - step); i ++) {if (!col[i]) {col[i] = 1;dfs_col (step + 1, i);col[i] = 0;}}
}void dfs_line (int step, int x) {//搜索每一行if (step > m2) {dfs_col (1, 0);return;}for (int i = x + 1; i <= m1 - (m2 - step); i ++) {if (!line[i]) {line[i] = 1;dfs_line (step + 1, i);line[i] = 0;}}
}int main () {scanf ("%d %d", &n1, &m1);for (int i = 1; i <= n1; i ++) {for (int j = 1; j <= m1; j ++) {scanf ("%d", &a[i][j]);}}scanf ("%d %d", &n2, &m2);for (int i = 1; i <= n2; i ++) {for (int j = 1; j <= m2; j ++) {scanf ("%d", &b[i][j]);}}dfs_line (1, 0);printf ("No");return 0;
}

D - “redocta”.swap(i,i+1)

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : 400400400 points

Problem Statement

You are given a string SSS that is a permutation of atcoder.
On this string S,S,S, you will perform the following operation 000 or more times:

Choose two adjacent characters of SSS and swap them.

Find the minimum number of operations required to make SSS equal atcoder.

Constraints

SSS is a string that is a permutation of atcoder

Input

Input is given from Standard Input in the following format:

SSS

Output

Print the answer as an integer.

题面翻译

给出字符串SSS，每次可以交换相邻的两个字母，求最少需要几次可以把SSS转化为atcoder。

Sample Input 1

catredo

Sample Output 1

You can make SSS equal atcoder in 888 operations as follows:
catredo → [ac]tredo → actre[od] → actr[oe]d → actro[de] → act[or]de → acto[dr]e → a[tc]odre → atcod[er]
This is the minimum number of operations achievable.

Sample Input 2

atcoder

Sample Output 2

In this case, the string S is already atcoder.

Sample Input 3

redocta

Sample Output 3

题解部分

这道题我们可以运用冒泡排序的思想，按照顺序，将指定字母交换至指定位置。由于atcoder不包含两个相同的字母，所以不用考虑最小的移动方法。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;int main () {int cnt = 0;string str = "atcoder", s;cin >> s;for (int i = 0; i < 7; i ++) {int p = s.find(str[i]);//寻找子串str[i]第一次出现的位置if (p < i) {//往后移(理论来说不需要判断)for (int j = p; j < i; j ++) {swap (s[j], s[j + 1]);cnt ++;}} else if (p > i) {//往前移for (int j = p; j > i; j --) {swap (s[j], s[j - 1]);cnt ++;}}}printf ("%d", cnt);return 0;
}

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freee Programming Contest 2022(AtCoder Beginner Contest 264) 题解 (A~D)

A - “atcoder”.substr()

Problem Statement

Constraints

Input

Output

题面翻译

Sample Input 1

Sample Output 1

Sample Input 2

Sample Output 2

Sample Input 3

Sample Output 3

题解部分

B - Nice Grid

Problem Statement

Constraints

Input

Output

题面翻译

Sample Input 1

Sample Output 1

Sample Input 2

Sample Output 2

题解部分

C - Matrix Reducing

Problem Statement

Constraints

Input

Output

题面翻译

Sample Input 1

Sample Output 1

Sample Input 2

Sample Output 2

题解部分

D - “redocta”.swap(i,i+1)

Problem Statement

Constraints

Input

Output

题面翻译

Sample Input 1

Sample Output 1

Sample Input 2

Sample Output 2

Sample Input 3

Sample Output 3

题解部分

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