2018 Multi-University Training Contest 1 记录

1001 Maximum Multiple

Problem Description

Given an integer n, Chiaki would like to find three positive integers x, y and z such that: $n=x+y+z$, $x\mid n$, $y \mid n$, $z \mid n$ and $xyz$ is maximum.

Input

There are multiple test cases. The first line of input contains an integer $T$ ($1 \le T \le 10^6$), indicating the number of test cases. For each test case:
The first line contains an integer $n$ ($1 \le n \le 10^{6}$).

Output

For each test case, output an integer denoting the maximum $xyz$. If there no such integers, output $-1$ instead.

Sample Input

3

1

2

3

Sample Output

-1

-1

1

思路

对 $n$ 是否是3,4的倍数进行讨论。

如果是 3 的倍数, 则 $ x = y = z = {n \over 3} $; 如果是 4 的倍数, 则 $ x = {n \over 2}, y = z = {n \over 4} $。

若 $n$ 为 3 和 4 的公倍数,则按 3 的倍数处理。

AC代码(队友)

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#include<bits/stdc++.h>
using namespace std;

#define LL long long

int n;

int main() {
int _;
scanf( "%d", &_ );
while( _-- ) {
scanf( "%d", &n );
if( n % 3 == 0 ) {
LL ans = 1;
ans = ( ans * n/3 * n/3 * n/3 );
printf( "%lld\n", ans );
} else if( n%4 == 0 ) {
LL ans = 1;
ans = ( ans * n/2 * n/4 * n/4 );
printf( "%lld\n", ans );

} else {
puts( "-1" );
}
}

return 0;
}

1011 Time Zone

Problem Description

Chiaki often participates in international competitive programming contests. The time zone becomes a big problem. Given a time in Beijing time (UTC +8), Chiaki would like to know the time in another time zone s.

Input

There are multiple test cases. The first line of input contains an integer $T$ ($1 \le T \le 10^6$), indicating the number of test cases. For each test case:
The first line contains two integers $a$, $b$ ($0 \le a \le 23, 0 \le b \le 59$) and a string $s$ in the format of "UTC+X’’, "UTC-X’’, "UTC+X.Y’’, or "UTC-X.Y’’ ($0 \le X, X.Y \le 14, 0 \le Y \le 9$).

Output

For each test, output the time in the format of $hh:mm$ (24-hour clock).

Sample Input

3

11 11 UTC+8

11 12 UTC+9

11 23 UTC+0

Sample Output

11:11

12:12

03:23

思路

水题,时区转换,模拟即可。

如果让自己写可能会以分钟计数,应该可以省去进位的麻烦。

AC代码(队友)

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#include<bits/stdc++.h>
using namespace std;

#define LL long long

int h, m;
double change;
char str[100];

int main() {
int _;
scanf( "%d", &_ );
while( _-- ) {
scanf( "%d%d", &h, &m );
getchar();
getchar();
getchar();
getchar();
scanf( "%lf", &change );
if( (int)change == change ) {
h += change-8;
if( h < 0 ) {
h += 24;
} else if( h >= 24 ) {
h -= 24;
}
} else {
int _h = (int)change;
double _m = change-_h;
//cout<<m<<endl;
if( _m < 0 ) {
_m = (int)( _m * 10 +- 0.5 ) / 10.0;
} else {
_m = (int)( _m * 10 + 0.5 ) / 10.0;
}

// cout<<_m<<endl;
h += _h-8;
m += (int)(_m*60);
// cout<<m<<endl;
if( m < 0 ) {
m += 60;
h -= 1;
} else if( m >= 60 ) {
m -= 60;
h += 1;
}
if( h < 0 ) {
h += 24;
} else if( h >= 24 ) {
h -= 24;
}
}
printf( "%02d:%02d\n", h, m );
}
}

1003 Triangle Partition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)

Special Judge

Problem Description

Chiaki has 3n points p1,p2,…,p3n. It is guaranteed that no three points are collinear. Chiaki would like to construct n disjoint triangles where each vertex comes from the 3n points.

Input

There are multiple test cases. The first line of input contains an integer $T$, indicating the number of test cases. For each test case:
The first line contains an integer $n$ ($1 \le n \le 1000$) – the number of triangle to construct.
Each of the next $3n$ lines contains two integers $x_i$ and $y_i$ ($-10^9 \le x_i, y_i \le 10^9$).
It is guaranteed that the sum of all $n$ does not exceed $10000$.

Output

For each test case, output $n$ lines contain three integers $a_i,b_i,c_i$ ($1 \le a_i,b_i,c_i \le 3n$) each denoting the indices of points the $i$-th triangle use. If there are multiple solutions, you can output any of them.

Sample Input

1

1

1 2

2 3

3 5

Sample Output

1 2 3

思路

把顶点按 x, y 坐标排序,依次构成三角形。

AC代码

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#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <utility>
#include <algorithm>
#define MAXN 3005
#define INF 0x3f3f3f3f
#define DEBUG
#define DataIn
typedef long long LL;

using namespace std;

int n;
struct point
{
int id, x, y;
bool operator < (point p)
{
if(x < p.x)
return true;
else if(x == p.x)
{
if(y < p.y)
return true;
else
return false;
}
else
return false;
}
}P[MAXN];

int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
for(int i=0; i<3*n; i++)
{
scanf("%d%d", &P[i].x, &P[i].y);
P[i].id = i+1;
}

sort(P, P+3*n);

for(int i=0; i<3*n; i++)
{
printf("%d ", P[i].id);
if(i+1%3==0)
putchar(10);
}
}
return 0;
}

1004 Distinct Values

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Chiaki has an array of $n$ positive integers. You are told some facts about the array: for every two elements $a_i$ and $a_j$ in the subarray $a_{l..r}$ ($l \le i \lt j \le r$), $a_i \ne a_j$ holds.
Chiaki would like to find a lexicographically minimal array which meets the facts.

Input

There are multiple test cases. The first line of input contains an integer $T$, indicating the number of test cases. For each test case:

The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^5$) – the length of the array and the number of facts. Each of the next $m$ lines contains two integers $l_i$ and $r_i$ ($1 \le l_i \le r_i \le n$).

It is guaranteed that neither the sum of all $n$ nor the sum of all $m$ exceeds $10^6$.

Output

For each test case, output $n$ integers denoting the lexicographically minimal array. Integers should be separated by a single space, and no extra spaces are allowed at the end of lines.

Sample Input

3

2 1

1 2

4 2

1 2

3 4

5 2

1 3

2 4

Samplt output

1 2

1 2 1 2

1 2 3 1 1

思路

用一个优先队列维护,优先使用较小的数字。使用完后注意放回。

AC代码(队友)

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#include<bits/stdc++.h>
using namespace std;

#define LL long long


const int N = 1e5+7;
struct node {
int l ,r;
} P[N];

int n, m;
int ans[N];

int cmp(node a,node b)
{
if(a.l!=b.r)
return a.l<b.l;
else
return a.r<b.r;
}

int l, r;

priority_queue<int, vector<int>, greater<int> > Q;

int main() {
int _;
scanf( "%d", &_ );
for( int i = 1; i <= 1e5; ++i ) {
Q.push( i );
}
while( _-- ) {
scanf( "%d%d", &n, &m );
for( int i = 0; i < m; ++i ) {
scanf( "%d%d", &P[i].l, &P[i].r );
}
sort( P, P+m, cmp );
l = r = 1;
ans[1] = 1;
Q.pop();

for( int i = 1; i <= n; ++i ) {
ans[i] = 0;
}
for( int i = 0; i < m; ++i ) {
int _l = P[i].l;
int _r = P[i].r;
for( int j = l; j < _l; ++j ) {
Q.push( ans[j] );
}
l = max( l, _l );
for( int j = r+1; j <= _r; ++j ) {
ans[j] = Q.top();
Q.pop();
}
r = max( r, _r );
}

for( int i = l; i <= r; ++i ) {
Q.push( ans[i] );
}
for( int i = 1; i <= n; ++i ) {
if( ans[i] == 0 ) {
ans[i] = 1;
}
if( i == 1 ) {
printf( "%d", ans[i] );
} else {
printf( " %d", ans[i] );
}
}
puts( "" );
}
}
❤采之欲遗谁,所思在远道。❤
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