AtCoder Beginner Contest 106 ABCD

A - Garden

Problem Statement

There is a farm whose length and width are A yard and B yard, respectively. A farmer, John, made a vertical road and a horizontal road inside the farm from one border to another, as shown below: (The gray part represents the roads.)

What is the area of this yard excluding the roads? Find it.

Note

It can be proved that the positions of the roads do not affect the area.

Constraints

• A is an integer between 2 and 100 (inclusive).
• B is an integer between 2 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

A B

Output

Print the area of this yard excluding the roads (in square yards).

Sample Input 1

Copy

2 2

Sample Output 1

Copy

1

In this case, the area is 1 square yard.

Sample Input 2

Copy

5 7

Sample Output 2

Copy

24

In this case, the area is 24 square yards.

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #define ll long long
 6 using namespace std;
 7
 8 int main(){
 9     int a,b;
10     cin >> a >> b;
11     printf("%d\n",a*b-(a+b-1));
12     return 0;
13 }

B - 105

Problem Statement

The number 105 is quite special - it is odd but still it has eight divisors. Now, your task is this: how many odd numbers with exactly eight positive divisors are there between 1 and N (inclusive)?

Constraints

• N is an integer between 1 and 200 (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print the count.

Sample Input 1

Copy

105

Sample Output 1

Copy

1

Among the numbers between 1 and 105, the only number that is odd and has exactly eight divisors is 105.

Sample Input 2

Copy

7

Sample Output 2

Copy

0

1 has one divisor. 3, 5 and 7 are all prime and have two divisors. Thus, there is no number that satisfies the condition.

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #define ll long long
 6 using namespace std;
 7
 8 int main(){
 9     int n;
10     cin >> n;
11     int ans = 0;
12     for(int i = 105; i <= n; i ++) {
13         if(i&1) {
14             int cnt = 0;
15             for(int j = 1; j <= n; j ++) if(i%j==0) cnt++;
16             if(cnt == 8) ans++;
17         }
18     }
19     cout << ans << endl;
20     return 0;
21 }

C - To Infinity

Problem Statement

Mr. Infinity has a string S consisting of digits from 1 to 9. Each time the date changes, this string changes as follows:

• Each occurrence of 2 in S is replaced with 22. Similarly, each 3 becomes 333, 4 becomes 4444, 5 becomes 55555, 6 becomes 666666, 7 becomes 7777777, 8 becomes 88888888 and 9 becomes 999999999. 1 remains as 1.

For example, if S is 1324, it becomes 1333224444 the next day, and it becomes 133333333322224444444444444444 the day after next. You are interested in what the string looks like after 5×1015 days. What is the K-th character from the left in the string after 5×1015 days?

Constraints

• S is a string of length between 1 and 100 (inclusive).
• K is an integer between 1 and 1018 (inclusive).
• The length of the string after 5×1015 days is at least K.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the K-th character from the left in Mr. Infinity‘s string after 5×1015 days.

Sample Input 1

Copy

1214
4

Sample Output 1

Copy

2

The string S changes as follows:

• Now: 1214
• After one day: 12214444
• After two days: 1222214444444444444444
• After three days: 12222222214444444444444444444444444444444444444444444444444444444444444444

The first five characters in the string after 5×1015 days is 12222. As K=4, we should print the fourth character, 2.

Sample Input 2

Copy

3
157

Sample Output 2

Copy

3

The initial string is 3. The string after 5×1015 days consists only of 3.

Sample Input 3

Copy

299792458
9460730472580800

Sample Output 3

Copy

2

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #define ll long long
 6 using namespace std;
 7 const int N = 110;
 8 char s[N];
 9 int main(){
10     ll k, l = 0;
11     cin >> s >> k;
12     ll len = strlen(s);
13     while(l < len && s[l] == ‘1‘) l ++;
14     if(l+1 <= k) printf("%c\n",s[l]);
15     else printf("1\n");
16     return 0;
17 }

D - AtCoder Express 2

Problem Statement

In Takahashi Kingdom, there is a east-west railroad and N cities along it, numbered 1, 2, 3, ..., N from west to east. A company called AtCoder Express possesses M trains, and the train i runs from City Li to City Ri (it is possible that Li=Ri). Takahashi the king is interested in the following Q matters:

• The number of the trains that runs strictly within the section from City pi to City qi, that is, the number of trains j such that pi≤Lj and Rj≤qi.

Although he is genius, this is too much data to process by himself. Find the answer for each of these Q queries to help him.

Constraints

• N is an integer between 1 and 500 (inclusive).
• M is an integer between 1 and 200 000 (inclusive).
• Q is an integer between 1 and 100 000 (inclusive).
• 1≤Li≤Ri≤N (1≤i≤M)
• 1≤pi≤qi≤N (1≤i≤Q)

Input

Input is given from Standard Input in the following format:

N M Q
L1 R1
L2 R2
:
LM RM
p1 q1
p2 q2
:
pQ qQ

Output

Print Q lines. The i-th line should contain the number of the trains that runs strictly within the section from City pi to City qi.

Sample Input 1

Copy

2 3 1
1 1
1 2
2 2
1 2

Sample Output 1

Copy

3

As all the trains runs within the section from City 1 to City 2, the answer to the only query is 3.

Sample Input 2

Copy

10 3 2
1 5
2 8
7 10
1 7
3 10

Sample Output 2

Copy

1
1

The first query is on the section from City 1 to 7. There is only one train that runs strictly within that section: Train 1. The second query is on the section from City 3 to 10. There is only one train that runs strictly within that section: Train 3.

Sample Input 3

Copy

10 10 10
1 6
2 9
4 5
4 7
4 7
5 8
6 6
6 7
7 9
10 10
1 8
1 9
1 10
2 8
2 9
2 10
3 8
3 9
3 10
1 10

Sample Output 3

Copy

7
9
10
6
8
9
6
7
8
10

可以用二维树状数组来做。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #define ll long long
 6 #define lowbit(x) x&(-x)
 7 using namespace std;
 8 const int MAX = 1010;
 9 int c[MAX][MAX];
10 void add(int x,int y,int k){
11     for(int i = x; i < MAX; i += lowbit(i)){
12         for(int j = y; j < MAX; j += lowbit(j)){
13             c[i][j] += k;
14         }
15     }
16 }
17 int query(int x, int y){
18     int sum = 0;
19     for(int i = x; i > 0; i -= lowbit(i)){
20         for(int j = y; j > 0; j -= lowbit(j)){
21             sum += c[i][j];
22         }
23     }
24     return sum;
25 }
26 int main(){
27     int n, m, q, l, r;
28     cin >> n >> m >> q;
29     for(int i = 1; i <= m; i ++) {
30         cin >> l >> r;
31         add(l,r,1);
32     }
33     while(q--) {
34         cin >> l >> r;
35         printf("%d\n",query(r,r)-query(l-1,r)-query(r,l-1)+query(l-1,l-1));
36     }
37     return 0;
38 }

原文地址：https://www.cnblogs.com/xingkongyihao/p/9501575.html