Educational Codeforces Round 7

A. Infinite Sequence

time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5…. The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).

Find the number on the n-th position of the sequence.
Input

The only line contains integer n (1 ≤ n ≤ 1014) — the position of the number to find.

Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output

Print the element in the n-th position of the sequence (the elements are numerated from one).
Sample test(s)
Input

3

Output

2

Input

5

Output

2

Input

10

Output

4

Input

55

Output

10

Input

56

Output

1

题意

1 1 2 1 2 3 的数列，问第n个位置是几

思路

直接从1开始减自然数，直到再减为负即为答案

代码

#include <stdio.h>
int main()
{
    long long n;
    scanf("%I64d",&n);
    long long i=1;
    while (n-i>0)
    {
        n=n-i;
        i++;
    }
    printf("%I64d",n);
} 

B. The Time

time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given the current time in 24-hour format hh:mm. Find and print the time after a minutes.

Note that you should find only the time after a minutes, see the examples to clarify the problem statement.

You can read more about 24-hour format here https://en.wikipedia.org/wiki/24-hour_clock.
Input

The first line contains the current time in the format hh:mm (0 ≤ hh < 24, 0 ≤ mm < 60). The hours and the minutes are given with two digits (the hours or the minutes less than 10 are given with the leading zeroes).

The second line contains integer a (0 ≤ a ≤ 104) — the number of the minutes passed.
Output

The only line should contain the time after a minutes in the format described in the input. Note that you should print exactly two digits for the hours and the minutes (add leading zeroes to the numbers if needed).

See the examples to check the input/output format.
Sample test(s)
Input

23:59
10

Output

00:09

Input

20:20
121

Output

22:21

Input

10:10
0

Output

10:10

题意

给出当前时间，问过n秒的时间是多少

思路

直接加就行了，注意处理前置0

代码

#include <stdio.h>
int main()
{
    int h,m,x;
    scanf("%d:%d %d",&h,&m,&x);
    h+=x/60;
    m+=x%60;
    if (m>=60)
    {
        h++;
        m-=60;
    }
    h=h%24;
    printf("%02d:%02d",h,m);
}

C. Not Equal on a Segment

time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi.

For the i-th query find any position pi (li ≤ pi ≤ ri) so that api ≠ xi.
Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the number of elements in a and the number of queries.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Each of the next m lines contains three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106) — the parameters of the i-th query.
Output

Print m lines. On the i-th line print integer pi — the position of any number not equal to xi in segment [li, ri] or the value  - 1 if there is no such number.
Sample test(s)
Input

6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2

Output

2
6
-1
4

题意

给一个串，有m个询问，问给定区间内是否有不等于x的

思路

直接处理每个询问结果TLE了，需要预处理。a[i]为给定串，b[i]等于第i个位置左侧第一个与a[i]不相等的数字的位置。那么对于每个询问，如果最右端点不相等，答案是最右端点，如果相等，考虑b[i]与左端点的关系就行了。

代码

#include <stdio.h>
int a[200010],b[200010];
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if (a[i]==a[i-1]) b[i]=b[i-1];
        else b[i]=i;
    }
    int l,r,x;
    while (m--)
    {
        int ans=0;
        scanf("%d %d %d",&l,&r,&x);
        if (a[r]!=x) ans=r;
        else if (b[r]>l) ans=b[r]-1;
        if (!ans) printf("-1\n");
        else printf("%d\n",ans);
    }
}

D. Optimal Number Permutation
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

You have array a that contains all integers from 1 to n twice. You can arbitrary permute any numbers in a.

Let number i be in positions xi, yi (xi < yi) in the permuted array a. Let’s define the value di = yi - xi — the distance between the positions of the number i. Permute the numbers in array a to minimize the value of the sum .
Input

The only line contains integer n (1 ≤ n ≤ 5·105).
Output
Print 2n integers — the permuted array a that minimizes the value of the sum s.
Sample test(s)
Input

2

Output

1 1 2 2

Input

1

Output

1 1

题意

从1到n，每个数用2次，要求得到的串的value和最小

思路

显然i间距为n-i时sum=0为最小，手动构造前几个观察规律就行
1 1
1 1 2 2
1 3 1 2 2 3
1 3 3 1 2 4 2 4
1 3 5 3 1 2 4 4 2 5

代码

#include <stdio.h>
int a[1000010];
int main()
{
    int n;
    scanf("%d",&n);
    int i;


    for (i=1;i<n;i+=2) printf("%d ",i);

    if (n%2) for (;i>=1;i-=2) printf("%d ",i);
    else for (i-=2;i>=1;i-=2) printf("%d ",i);

    for (i=2;i<=n;i+=2) printf("%d ",i);

    if (n%2) for (i-=2;i>=2;i-=2) printf("%d ",i);
    else for (i-=4;i>=2;i-=2) printf("%d ",i);
    printf("%d",n);
}

E,F不太会。。。。