AtCoder Beginner Contest 103
https://beta.atcoder.jp/contests/abc103
A - Task Scheduling Problem
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 100100 points
Problem Statement
You have three tasks, all of which need to be completed.
First, you can complete any one task at cost 00.
Then, just after completing the ii-th task, you can complete the jj-th task at cost |Aj−Ai||Aj−Ai|.
Here, |x||x| denotes the absolute value of xx.
Find the minimum total cost required to complete all the task.
Constraints
- All values in input are integers.
- 1≤A1,A2,A3≤1001≤A1,A2,A3≤100
Input
Input is given from Standard Input in the following format:
A1A1 A2A2 A3A3
Output
Print the minimum total cost required to complete all the task.
Sample Input 1 Copy
1 6 3
Sample Output 1 Copy
5
When the tasks are completed in the following order, the total cost will be 55, which is the minimum:
- Complete the first task at cost 00.
- Complete the third task at cost 22.
- Complete the second task at cost 33.
Sample Input 2 Copy
11 5 5
Sample Output 2 Copy
6
Sample Input 3 Copy
100 100 100
Sample Output 3 Copy
0 最大的减去最小的。
import java.util.Arrays;
import java.util.Scanner;
import static java.lang.Math.*;
public class Main { public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int [] a = new int[3];
for(int i = 0;i < 3;i ++) {
a[i] = in.nextInt();
}
Arrays.sort(a);
System.out.println(a[2] - a[0]);
}
}
B - String Rotation
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 200200 points
Problem Statement
You are given string SS and TT consisting of lowercase English letters.
Determine if SS equals TT after rotation.
That is, determine if SS equals TT after the following operation is performed some number of times:
Operation: Let S=S1S2...S|S|S=S1S2...S|S|. Change SS to S|S|S1S2...S|S|−1S|S|S1S2...S|S|−1.
Here, |X||X| denotes the length of the string XX.
Constraints
- 2≤|S|≤1002≤|S|≤100
- |S|=|T||S|=|T|
- SS and TT consist of lowercase English letters.
Input
Input is given from Standard Input in the following format:
SS
TT
Output
If SS equals TT after rotation, print Yes; if it does not, print No.
Sample Input 1 Copy
kyoto
tokyo
Sample Output 1 Copy
Yes
- In the first operation,
kyotobecomesokyot. - In the second operation,
okyotbecomestokyo.
Sample Input 2 Copy
abc
arc
Sample Output 2 Copy
No
abc does not equal arc after any number of operations.
Sample Input 3 Copy
aaaaaaaaaaaaaaab
aaaaaaaaaaaaaaab
Sample Output 3 Copy
Yes
排着遍历拼接。
import java.util.Arrays;
import java.util.Scanner;
import static java.lang.Math.*;
public class Main {
static boolean check(String a,String b) {
if(a.length() != b.length())return false;
for(int i = 0;i < a.length();i ++) {
if(a.charAt(i) == b.charAt(0) && (a.substring(i, a.length()) + a.substring(0, i)).equals(b)) {
return true;
}
}
return false;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String a = in.nextLine();
String b = in.nextLine();
System.out.println(check(a,b) ? "Yes" : "No");
}
}
C - Modulo Summation
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 300300 points
Problem Statement
You are given NN positive integers a1,a2,...,aNa1,a2,...,aN.
For a non-negative integer mm, let f(m)=(m mod a1)+(m mod a2)+...+(m mod aN)f(m)=(m mod a1)+(m mod a2)+...+(m mod aN).
Here, X mod YX mod Y denotes the remainder of the division of XX by YY.
Find the maximum value of ff.
Constraints
- All values in input are integers.
- 2≤N≤30002≤N≤3000
- 2≤ai≤1052≤ai≤105
Input
Input is given from Standard Input in the following format:
NN
a1a1 a2a2 ...... aNaN
Output
Print the maximum value of ff.
Sample Input 1 Copy
3
3 4 6
Sample Output 1 Copy
10
f(11)=(11 mod 3)+(11 mod 4)+(11 mod 6)=10f(11)=(11 mod 3)+(11 mod 4)+(11 mod 6)=10 is the maximum value of ff.
Sample Input 2 Copy
5
7 46 11 20 11
Sample Output 2 Copy
90
Sample Input 3 Copy
7
994 518 941 851 647 2 581
Sample Output 3 Copy
4527
一个数对多个数去摸和要最大。
对于m,m%a1最大值是a1 - 1,也就是说令m = t1 * a1 - 1(t1 >= 1)时m%a1最大,同样如果对于a1~n都满足m = ti * ai - 1,总和就会最大,实际上就是满足m = t1 * a1 - 1 = t2 * a2 - 1 = ...
= tn * an - 1,即t1 * a1 = t2 * a2 = ... = tn * an = m + 1,所以m + 1是a1~n的公倍数,是存在的。
答案就显而易见是所有属的和-n。
import java.util.Arrays;
import java.util.Scanner;
import static java.lang.Math.*;
public class Main {
static boolean check(String a,String b) {
if(a.length() != b.length())return false;
for(int i = 0;i < a.length();i ++) {
if(a.charAt(i) == b.charAt(0) && (a.substring(i, a.length()) + a.substring(0, i)).equals(b)) {
return true;
}
}
return false;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int d;
int ans = -n;
for(int i = 0;i < n;i ++) {
d = in.nextInt();
ans += d;
}
System.out.println(ans);
}
}
D - Islands War
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 400400 points
Problem Statement
There are NN islands lining up from west to east, connected by N−1N−1 bridges.
The ii-th bridge connects the ii-th island from the west and the (i+1)(i+1)-th island from the west.
One day, disputes took place between some islands, and there were MM requests from the inhabitants of the islands:
Request ii: A dispute took place between the aiai-th island from the west and the bibi-th island from the west. Please make traveling between these islands with bridges impossible.
You decided to remove some bridges to meet all these MM requests.
Find the minimum number of bridges that must be removed.
Constraints
- All values in input are integers.
- 2≤N≤1052≤N≤105
- 1≤M≤1051≤M≤105
- 1≤ai<bi≤N1≤ai<bi≤N
- All pairs (ai,bi)(ai,bi) are distinct.
Input
Input is given from Standard Input in the following format:
NN MM
a1a1 b1b1
a2a2 b2b2
::
aMaM bMbM
Output
Print the minimum number of bridges that must be removed.
Sample Input 1 Copy
5 2
1 4
2 5
Sample Output 1 Copy
1
The requests can be met by removing the bridge connecting the second and third islands from the west.
Sample Input 2 Copy
9 5
1 8
2 7
3 5
4 6
7 9
Sample Output 2 Copy
2
Sample Input 3 Copy
5 10
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
Sample Output 3 Copy
4
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
import static java.lang.Math.*;
class bridge{
public int a;
public int b;
public bridge() {
a = b = 0;
}
}
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
bridge [] s = new bridge[k];
for(int i = 0;i < k;i ++) {
s[i] = new bridge();
s[i].a = in.nextInt();
s[i].b = in.nextInt();
}
Arrays.sort(s,new Comparator<bridge>() {
public int compare(bridge x,bridge y) {
if(x.a == x.b)return x.b - y.b;
return x.a - y.a;
}
});
int ans = 0,r = 0;
for(int i = 0;i < k;i ++) {
if(s[i].a >= r) {
ans ++;
r = s[i].b;
}
else if(r > s[i].b)r = s[i].b;
}
System.out.println(ans);
}
}
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