Codeforces Round #615 (Div. 3) 题解

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本次打的极差，我哭了！！！

A - Collecting Coins

题目

A. Collecting Coins

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp has three sisters: Alice, Barbara, and Cerene. They’re collecting coins. Currently, Alice has aa coins, Barbara has bb coins and Cerene has cc coins. Recently Polycarp has returned from the trip around the world and brought nn coins.

He wants to distribute all these nn coins between his sisters in such a way that the number of coins Alice has is equal to the number of coins Barbara has and is equal to the number of coins Cerene has. In other words, if Polycarp gives AA coins to Alice, BB coins to Barbara and CC coins to Cerene (A+B+C=nA+B+C=n), then a+A=b+B=c+Ca+A=b+B=c+C.

Note that A, B or C (the number of coins Polycarp gives to Alice, Barbara and Cerene correspondingly) can be 0.

Your task is to find out if it is possible to distribute all nn coins between sisters in a way described above.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The next tt lines describe test cases. Each test case is given on a new line and consists of four space-separated integers a,b,ca,b,c and nn (1≤a,b,c,n≤1081≤a,b,c,n≤108) — the number of coins Alice has, the number of coins Barbara has, the number of coins Cerene has and the number of coins Polycarp has.

Output

For each test case, print “YES” if Polycarp can distribute all nn coins between his sisters and “NO” otherwise.

Example

input

Copy

5
5 3 2 8
100 101 102 105
3 2 1 100000000
10 20 15 14
101 101 101 3

output

Copy

YES
YES
NO
NO
YES

题意

给你四个数a，b，c，d，n.问你是否能将n拆成三个数A,B,C，使得A+a=B+b=C+c。

思路

先计算三个数的差值的绝对值abs,如果abs大于n则肯定不行，如果小于n，还需判断(n-abs)%3是否为0，不为0则不行。

代码

import sys
import queue
sys.setrecursionlimit(10 ** 9)
IA = lambda: map(int, input().split())
IAL = lambda: list(map(int, input().split()))
IM = lambda N: [IA() for _ in range(N)]
# !/usr/bin/env python3
# -*- coding: utf-8 -*-
N = 105
T=int(input())
for i in range(0,T):a,b,c,n=IA()#print(n-(2*max(a,b,c)-(a+b+c-max(a,b,c))))if (2*max(a,b,c)-(a+b+c-max(a,b,c)))>n:print("NO")elif (n-(2*max(a,b,c)-(a+b+c-max(a,b,c))))%3==0:print("YES")else:print("NO")

B. Collecting Packages

题目

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a robot in a warehouse and nn packages he wants to collect. The warehouse can be represented as a coordinate grid. Initially, the robot stays at the point (0,0)(0,0). The ii-th package is at the point (xi,yi)(xi,yi). It is guaranteed that there are no two packages at the same point. It is also guaranteed that the point (0,0)(0,0) doesn’t contain a package.

The robot is semi-broken and only can move up (‘U’) and right (‘R’). In other words, in one move the robot can go from the point (x,y)(x,y) to the point (x+1,yx+1,y) or to the point (x,y+1)(x,y+1).

As we say above, the robot wants to collect all nn packages (in arbitrary order). He wants to do it with the minimum possible number of moves. If there are several possible traversals, the robot wants to choose the lexicographically smallest path.

The string ss of length nn is lexicographically less than the string tt of length nn if there is some index 1≤j≤n1≤j≤n that for all ii from 11 to j−1j−1 si=tisi=ti and sj<tjsj<tj. It is the standard comparison of string, like in a dictionary. Most programming languages compare strings in this way.

Input

The first line of the input contains an integer tt (1≤t≤1001≤t≤100) — the number of test cases. Then test cases follow.

The first line of a test case contains one integer nn (1≤n≤10001≤n≤1000) — the number of packages.

The next nn lines contain descriptions of packages. The ii-th package is given as two integers xixi and yiyi (0≤xi,yi≤10000≤xi,yi≤1000) — the xx-coordinate of the package and the yy-coordinate of the package.

It is guaranteed that there are no two packages at the same point. It is also guaranteed that the point (0,0)(0,0) doesn’t contain a package.

The sum of all values nn over test cases in the test doesn’t exceed 10001000.

Output

Print the answer for each test case.

If it is impossible to collect all nn packages in some order starting from (0,00,0), print “NO” on the first line.

Otherwise, print “YES” in the first line. Then print the shortest path — a string consisting of characters ‘R’ and ‘U’. Among all such paths choose the lexicographically smallest path.

Note that in this problem “YES” and “NO” can be only uppercase words, i.e. “Yes”, “no” and “YeS” are not acceptable.

Example

input

Copy

output

Copy

YES
RUUURRRRUU
NO
YES
RRRRUUU

Note

For the first test case in the example the optimal path RUUURRRRUU is shown below:

题意

给出直角坐标系中的一些坐标，你只能向上(‘U’)或向右(‘R’)走，问你能否走完这些坐标，如果能请输出字典序最小的行进路线

思路

先x后y坐标排序，如果有一个点的纵坐标在上一个点的纵坐标下方，则无法走。输出顺序只需要从上一个点先走R再走U走到下一个点即可

代码

import sys
sys.setrecursionlimit(10**9)
IA =lambda: map(int,input().split())T=int(input())
for t in range(0,T):n=int(input())a=[]for i in range(0,n):x,y=IA()a.append([x,y])a.sort()sx=0sy=0ans=""flag=1for item in a:if item[0]>=sx and item[1]>=sy:ans=ans+(item[0]-sx)*'R'+(item[1]-sy)*'U'sx=item[0]sy=item[1]else:print("NO")flag=-1breakif flag==1:print("YES")print(ans)

C - Product of Three Numbers

题目

C. Product of Three Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given one integer number nn. Find three distinct integers a,b,ca,b,c such that 2≤a,b,c2≤a,b,c and a⋅b⋅c=na⋅b⋅c=n or say that it is impossible to do it.

If there are several answers, you can print any.

You have to answer tt independent test cases.

Input

The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases.

The next nn lines describe test cases. The ii-th test case is given on a new line as one integer nn (2≤n≤1092≤n≤109).

Output

For each test case, print the answer on it. Print “NO” if it is impossible to represent nn as a⋅b⋅ca⋅b⋅c for some distinct integers a,b,ca,b,c such that 2≤a,b,c2≤a,b,c.

Otherwise, print “YES” and any possible such representation.

Example

input

Copy

output

Copy

YES
2 4 8 
NO
NO
NO
YES
3 5 823

题意

给你一个数n，问你是否能将n分解成三个不同的数相乘，可以的话则输出任一解

思路

只需要将遍历从2到，如果i为n的因子，就将i记录，并将n除以i。这样的操作执行三次之后，就可以跳出循环。

判断条件直接暴力就行，因为就三个。

代码

import sys
sys.setrecursionlimit(10**9)
IA =lambda: map(int,input().split())
ans=[0,0,0,0]
def solve(n):num=int(0)i=int(2)tmp=1m=nwhile i*i<=n:if n%i==0:n=n//itmp*=iif tmp not in ans:num+=1ans[num]=tmptmp=1if num>=3:ans[num]*=nreturn 1i+=1if n>1:if tmp*n not in ans:num+=1ans[num]=n*tmpif num>=3: return 1else:return -1T=int(input())
for t in range(0,T):ans=[0,0,0,0]n=int(input())if solve(n)==1:print("YES")for i in range(1,4):print(ans[i],end=" ")print()else:print("NO")

D. MEX maximizing

题目

D. MEX maximizing

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Recall that MEX of an array is a minimum non-negative integer that does not belong to the array. Examples:

for the array [0,0,1,0,2][0,0,1,0,2] MEX equals to 33 because numbers 0,10,1 and 22 are presented in the array and 33 is the minimum non-negative integer not presented in the array;
for the array [1,2,3,4][1,2,3,4] MEX equals to 00 because 00 is the minimum non-negative integer not presented in the array;
for the array [0,1,4,3][0,1,4,3] MEX equals to 22 because 22 is the minimum non-negative integer not presented in the array.

You are given an empty array a=[]a=[] (in other words, a zero-length array). You are also given a positive integer xx.

You are also given qq queries. The jj-th query consists of one integer yjyj and means that you have to append one element yjyj to the array. The array length increases by 11 after a query.

In one move, you can choose any index ii and set ai:=ai+xai:=ai+x or ai:=ai−xai:=ai−x (i.e. increase or decrease any element of the array by xx). The only restriction is that aiai cannot become negative. Since initially the array is empty, you can perform moves only after the first query.

You have to maximize the MEX (minimum excluded) of the array if you can perform any number of such operations (you can even perform the operation multiple times with one element).

You have to find the answer after each of qq queries (i.e. the jj-th answer corresponds to the array of length jj).

Operations are discarded before each query. I.e. the array aa after the jj-th query equals to [y1,y2,…,yj][y1,y2,…,yj].

Input

The first line of the input contains two integers q,xq,x (1≤q,x≤4⋅1051≤q,x≤4⋅105) — the number of queries and the value of xx.

The next qq lines describe queries. The jj-th query consists of one integer yjyj (0≤yj≤1090≤yj≤109) and means that you have to append one element yjyj to the array.

Output

Print the answer to the initial problem after each query — for the query jj print the maximum value of MEX after first jj queries. Note that queries are dependent (the array changes after each query) but operations are independent between queries.

Examples

input

Copy

output

Copy

input

Copy

output

Copy

Note

In the first example:

After the first query, the array is a=[0]a=[0]: you don’t need to perform any operations, maximum possible MEX is 11.
After the second query, the array is a=[0,1]a=[0,1]: you don’t need to perform any operations, maximum possible MEX is 22.
After the third query, the array is a=[0,1,2]a=[0,1,2]: you don’t need to perform any operations, maximum possible MEX is 33.
After the fourth query, the array is a=[0,1,2,2]a=[0,1,2,2]: you don’t need to perform any operations, maximum possible MEX is 33 (you can’t make it greater with operations).
After the fifth query, the array is a=[0,1,2,2,0]a=[0,1,2,2,0]: you can perform a[4]:=a[4]+3=3a[4]:=a[4]+3=3. The array changes to be a=[0,1,2,2,3]a=[0,1,2,2,3]. Now MEX is maximum possible and equals to 44.
After the sixth query, the array is a=[0,1,2,2,0,0]a=[0,1,2,2,0,0]: you can perform a[4]:=a[4]+3=0+3=3a[4]:=a[4]+3=0+3=3. The array changes to be a=[0,1,2,2,3,0]a=[0,1,2,2,3,0]. Now MEX is maximum possible and equals to 44.
After the seventh query, the array is

a=[0,1,2,2,0,0,10]a=[0,1,2,2,0,0,10]

. You can perform the following operations:
- a[3]:=a[3]+3=2+3=5a[3]:=a[3]+3=2+3=5,
- a[4]:=a[4]+3=0+3=3a[4]:=a[4]+3=0+3=3,
- a[5]:=a[5]+3=0+3=3a[5]:=a[5]+3=0+3=3,
- a[5]:=a[5]+3=3+3=6a[5]:=a[5]+3=3+3=6,
- a[6]:=a[6]−3=10−3=7a[6]:=a[6]−3=10−3=7,
- a[6]:=a[6]−3=7−3=4a[6]:=a[6]−3=7−3=4.
The resulting array will be

a=[0,1,2,5,3,6,4]a=[0,1,2,5,3,6,4]

. Now MEX is maximum possible and equals to 7

题意
- 输入q，x，然后输入q个数（q个询问)，每次放进一个数（开始为空数组）
- 同时每次可以对数组内的某个数或者某几个数做任意操作数的加x或者减x （x为input第二个数），
- 输出：处理完后询问，在该数组情况下的最小整数（就是数组里的数都被排除的情况下）从0开始找，第一个不在数组里的整数就是了，然后要求操作加x或者减x的任意操作后使得那个询问结果最大化。
思路

构建mod x的剩余系，每次找到这个剩余系里面最大的加上x即可，从0开始判断即可，直接看代码，很短。

代码
```
import sys
sys.setrecursionlimit(10**9)
from collections import defaultdictIA =lambda: map(int,input().split())
ans=[0,0,0,0]T,x=IA()
se=defaultdict(int)
be=0for t in range(0,T):a=int(input())temp=0se[a%x]+=1while se[be%x] > 0:se[be%x]-=1be+=1print(be)
```