Title: interval number even number
Xiao Ming these days have been thinking about such a strange and interesting question:
How many even numbers in the range of 1 to do a full array of N? Here's the definition of even number range is:
If the interval [L, R] in all elements (i.e., in this arrangement the L-th to the R elements)
after ascending order to get a length of "continuous" series R-L + 1, and called this interval even number range.
When N was very young, Xiao Ming can quickly calculate the answer, but when N becomes large,
the problem is not so simple, and now Bob needs your help.
Input format:
The first line is a positive integer N (1 <= N <= 50000), represents the size of the whole arrangement.
The second line is the N different number Pi (1 <= Pi <= N), which represents a certain N digital full array.
Output Format:
Output an integer representing the number of different sections even numbers.
Example:
User input:
. 4
. 3. 1 2. 4
Program should output:
7
The user input:
. 5
. 3. 4. 5. 1 2
Program should output:
9
Explanation:
The first use embodiment, seven sections even numbers are: [1,1], [1,2], [1,3], [1,4], [2,2], [3, 3], [4,4]
second use case, there are nine sections even numbers are: [1,1], [1,2], [1,3], [1,4], [1, 5], [2,2], [3,3], [4,4], [5,5]
Resources for:
peak memory consumption <64M
the CPU consumption <5000ms
Please strictly follow the requirements of output, not superfluous to print something like: "Please enter ..." unwanted content.
All source code in a single file, through debugging, submit the copy source.
Note: main function needs to return 0
Note: Use only ANSI C / ANSI C ++ standard, do not call a special function depends on the build environment or operating system.
Note: all dependent functions must explicitly #include <xxx>, can not be provided by the project source file header files commonly omitted.
When you submit, pay attention to select the desired type of compiler.
solution:
#include <iostream>
using namespace std;
int n;
int arr[50000];
int ans;
int main(int argc, const char *argv[]) {
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", &arr[i]);
}
for (int j = 0; j <= n - 1; ++j) {
int min=arr[j];
int max = arr[j];
for (int i = j; i <= n - 1; ++i) {
if(arr[i]>max)
max=arr[i];
if(arr[i]<min)
min=arr[i];
if (i == j)
ans++;
else {
if(max-min+1==i-j+1)
ans++;
}
}
}
printf("%d\n",ans);
return 0;
}Official Solution:
#include <iostream>
using namespace std;
int n;
int arr[50000];
int ans;
int main(int argc, const char *argv[]) {
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", &arr[i]);
}
for (int j = 0; j <= n - 1; ++j) {
int min=arr[j];
int max = arr[j];
for (int i = j; i <= n - 1; ++i) {
if(arr[i]>max)
max=arr[i];
if(arr[i]<min)
min=arr[i];
if (i == j)
ans++;
else {
if(max-min+1==i-j+1)
ans++;
}
}
}
printf("%d\n",ans);
return 0;
}