A - Multiple of 2 and N
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 100100 points
Problem Statement
You are given a positive integer NN. Find the minimum positive integer divisible by both 22 and NN.
Constraints
- 1≤N≤1091≤N≤109
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
NN
Output
Print the minimum positive integer divisible by both 22 and NN.
Sample Input 1 Copy
Copy
3
Sample Output 1 Copy
Copy
6
66 is divisible by both 22 and 33. Also, there is no positive integer less than 66 that is divisible by both 22 and 33. Thus, the answer is 66.
Sample Input 2 Copy
Copy
10
Sample Output 2 Copy
Copy
10
Sample Input 3 Copy
Copy
999999999
Sample Output 3 Copy
Copy
1999999998代码：
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
if(n % 2 == 1)n *= 2;
System.out.println(n);
}
}
B - Maximum Difference
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 200200 points
Problem Statement
You are given an integer sequence AA of length NN. Find the maximum absolute difference of two elements (with different indices) in AA.
Constraints
- 2≤N≤1002≤N≤100
- 1≤Ai≤1091≤Ai≤109
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
NN A1A1 A2A2 ...... ANAN
Output
Print the maximum absolute difference of two elements (with different indices) in AA.
Sample Input 1 Copy
Copy
4 1 4 6 3
Sample Output 1 Copy
Copy
5
The maximum absolute difference of two elements is A3−A1=6−1=5A3−A1=6−1=5.
Sample Input 2 Copy
Copy
2 1000000000 1
Sample Output 2 Copy
Copy
999999999
Sample Input 3 Copy
Copy
5 1 1 1 1 1
Sample Output 3 Copy
Copy
0代码：
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int max = 0,min = 1000000000;
for(int i = 0;i < n;i ++) {
int d = in.nextInt();
max = Math.max(max, d);
min = Math.min(min, d);
}
System.out.println(max - min);
}
}
C - Linear Approximation
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 300300 points
Problem Statement
Snuke has an integer sequence AA of length NN.
He will freely choose an integer bb. Here, he will get sad if AiAi and b+ib+i are far from each other. More specifically, the sadness of Snuke is calculated as follows:
- abs(A1−(b+1))+abs(A2−(b+2))+...+abs(AN−(b+N))abs(A1−(b+1))+abs(A2−(b+2))+...+abs(AN−(b+N))
Here, abs(x)abs(x) is a function that returns the absolute value of xx.
Find the minimum possible sadness of Snuke.
Constraints
- 1≤N≤2×1051≤N≤2×105
- 1≤Ai≤1091≤Ai≤109
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
NN A1A1 A2A2 ...... ANAN
Output
Print the minimum possible sadness of Snuke.
Sample Input 1 Copy
Copy
5 2 2 3 5 5
Sample Output 1 Copy
Copy
2
If we choose b=0b=0, the sadness of Snuke would be abs(2−(0+1))+abs(2−(0+2))+abs(3−(0+3))+abs(5−(0+4))+abs(5−(0+5))=2abs(2−(0+1))+abs(2−(0+2))+abs(3−(0+3))+abs(5−(0+4))+abs(5−(0+5))=2. Any choice of bb does not make the sadness of Snuke less than 22, so the answer is 22.
Sample Input 2 Copy
Copy
9 1 2 3 4 5 6 7 8 9
Sample Output 2 Copy
Copy
0
Sample Input 3 Copy
Copy
6 6 5 4 3 2 1
Sample Output 3 Copy
Copy
18
Sample Input 4 Copy
Copy
7 1 1 1 1 2 3 4
Sample Output 4 Copy
Copy
6先把所有的值减去对应下标的值，之后的序列是求都减去同一个值后，绝对值的和最小，那么就排序，找到中间位置的值，所有的数减去中间位置的值就可以了，中间位置的左边比他小，右边比他大，减去值后中间的变为0，左边的都增加，右边的豆腐减少，保证减少的小于等于增加的即可。代码：
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int [] s = new int[n];
long d = 0;
for(int i = 0;i < n;i ++) {
s[i] = in.nextInt();
s[i] -= i + 1;
}
Arrays.sort(s);
for(int i = 0;i < n;i ++) {
d += Math.abs(s[i] - s[n / 2]);
}
System.out.println(d);
}
}
D - Equal Cut
Time Limit: 2 sec / Memory Limit: 1024 MB
Score : 600600 points
Problem Statement
Snuke has an integer sequence AA of length NN.
He will make three cuts in AA and divide it into four (non-empty) contiguous subsequences B,C,DB,C,D and EE. The positions of the cuts can be freely chosen.
Let P,Q,R,SP,Q,R,S be the sums of the elements in B,C,D,EB,C,D,E, respectively. Snuke is happier when the absolute difference of the maximum and the minimum among P,Q,R,SP,Q,R,S is smaller. Find the minimum possible absolute difference of the maximum and the minimum among P,Q,R,SP,Q,R,S.
Constraints
- 4≤N≤2×1054≤N≤2×105
- 1≤Ai≤1091≤Ai≤109
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
NN A1A1 A2A2 ...... ANAN
Output
Find the minimum possible absolute difference of the maximum and the minimum among P,Q,R,SP,Q,R,S.
Sample Input 1 Copy
Copy
5 3 2 4 1 2
Sample Output 1 Copy
Copy
2
If we divide AA as B,C,D,E=(3),(2),(4),(1,2)B,C,D,E=(3),(2),(4),(1,2), then P=3,Q=2,R=4,S=1+2=3P=3,Q=2,R=4,S=1+2=3. Here, the maximum and the minimum among P,Q,R,SP,Q,R,Sare 44 and 22, with the absolute difference of 22. We cannot make the absolute difference of the maximum and the minimum less than 22, so the answer is 22.
Sample Input 2 Copy
Copy
10 10 71 84 33 6 47 23 25 52 64
Sample Output 2 Copy
Copy
36
Sample Input 3 Copy
Copy
7 1 2 3 1000000000 4 5 6
Sample Output 3 Copy
Copy
999999994如果是排着去找这三个位置，是O(n^3)的，可以采用折半枚举的技巧，先找到第二个位置，然后两边各自分为两块，用l和r标记，两块的差距尽可能小即可，随着i的右移，lr只需要初始化一次，然后跟着移动即可，i的移动必然会打破原来的平衡，lr只能继续右移，不需要从最初端开始，不然仍然超时。java代码：
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
long [] s = new long[n + 1];
for(int i = 1;i <= n;i ++) {
s[i] = in.nextLong();
s[i] += s[i - 1];
}
int l = 1,r = 3;
long ans = s[n];
for(int i = 2;i < n - 1;i ++) {
long last = s[n];
while(l < i) {
long d = Math.abs(s[i] - s[l] * 2);
if(d > last) break;
else last = d;
l ++;
}
l --;
last = s[n];
while(r < n) {
long d = Math.abs(s[n] + s[i] - s[r] * 2);
if(d > last) break;
else last = d;
r ++;
}
r --;
long max = Math.max(Math.max(Math.max(s[l], s[i] - s[l]), s[r] - s[i]), s[n] - s[r]);
long min = Math.min(Math.min(Math.min(s[l], s[i] - s[l]), s[r] - s[i]), s[n] - s[r]);
ans = Math.min(max - min, ans);
}
System.out.println(ans);
}
}
c++代码：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define Max 200005
using namespace std;
typedef long long LL;
int main() {
int n;
int a[Max];
LL sum[Max] = {0};
scanf("%d",&n);
for(int i = 1;i <= n;i ++) {
scanf("%d",&a[i]);
sum[i] = sum[i - 1] + a[i];
}
LL ans = sum[n];
int l = 1,r = 3;
for(int i = 2;i <= n - 2;i ++) {
LL last = sum[n];
while(l < i) {
LL d = abs(sum[i] - sum[l] * 2);
if(d > last) break;
else last = d;
l ++;
}
l --;
last = sum[n];
while(r < n) {
LL d = abs(sum[n] + sum[i] - sum[r] * 2);
if(d > last) break;
else last = d;
r ++;
}
r --;
LL up = max(max(max(sum[l],sum[i] - sum[l]),sum[r] - sum[i]),sum[n] - sum[r]);
LL down = min(min(min(sum[l],sum[i] - sum[l]),sum[r] - sum[i]),sum[n] - sum[r]);
ans = min(ans,up - down);
}
printf("%lld",ans);
}
原文地址：https://www.cnblogs.com/8023spz/p/9452315.html