http://www.yyycode.cn/index.php/2020/05/30/codeforces-round-645-div-2/
A. Park Lightingtime
limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output
Due to the coronavirus pandemic, city authorities obligated citizens to keep a social distance. The mayor of the city Semyon wants to light up Gluharniki park so that people could see each other even at night to keep the social distance.
The park is a rectangular table with nn rows and mm columns, where the cells of the table are squares, and the boundaries between the cells are streets. External borders are also streets. Every street has length 11. For example, park with n=m=2n=m=2 has 1212 streets.
You were assigned to develop a plan for lighting the park. You can put lanterns in the middle of the streets. The lamp lights two squares near it (or only one square if it stands on the border of the park).
The park sizes are: n=4n=4, m=5m=5. The lighted squares are marked yellow. Please note that all streets have length 11. Lanterns are placed in the middle of the streets. In the picture not all the squares are lit.
Semyon wants to spend the least possible amount of money on lighting but also wants people throughout the park to keep a social distance. So he asks you to find the minimum number of lanterns that are required to light all the squares.Input
The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases in the input. Then tt test cases follow.
Each test case is a line containing two integers nn, mm (1≤n,m≤1041≤n,m≤104) — park sizes.Output
Print tt answers to the test cases. Each answer must be a single integer — the minimum number of lanterns that are required to light all the squares.ExampleinputCopy
5 1 1 1 3 2 2 3 3 5 3
outputCopy
1 2 2 5 8
Note
Possible optimal arrangement of the lanterns for the 22-nd test case of input data example:
Possible optimal arrangement of the lanterns for the 33-rd test case of input data example:
題意：給一個n,m爲邊的矩形，問最少的燈使得整個矩形照亮
思路：又考了奇數偶數（話說奇偶性考得真的多
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5;
typedef long long LL;
int main(void)
{
LL t;cin>>t;
while(t--)
{
LL n,m;cin>>n>>m;
if(n==1&&m==1) cout<<1<<endl;
else
{
if(n%2==0&&m%2==0)
{
cout<<(m/2)*n<<endl;
}
else if(n%2==1&&m%2==0)
{
cout<<(m/2)*n<<endl;
}
else if(n%2==0&&m%2==1)
{
cout<<(n/2)*m<<endl;
}
else if(n%2==1&&m%2==1)
{
LL k=(m-1)/2*n;
k+=(n+1)/2;
cout<<k<<endl;
}
}
}
return 0;
}
B. Maria Breaks the Self-isolation
Maria is the most active old lady in her house. She was tired of sitting at home. She decided to organize a ceremony against the coronavirus.
She has nn friends who are also grannies (Maria is not included in this number). The ii-th granny is ready to attend the ceremony, provided that at the time of her appearance in the courtyard there will be at least aiai other grannies there. Note that grannies can come into the courtyard at the same time. Formally, the granny ii agrees to come if the number of other grannies who came earlier or at the same time with her is greater than or equal to aiai.
Grannies gather in the courtyard like that.
- Initially, only Maria is in the courtyard (that is, the initial number of grannies in the courtyard is 11). All the remaining nn grannies are still sitting at home.
- On each step Maria selects a subset of grannies, none of whom have yet to enter the courtyard. She promises each of them that at the time of her appearance there will be at least aiai other grannies (including Maria) in the courtyard. Maria can call several grannies at once. In this case, the selected grannies will go out into the courtyard at the same moment of time.
- She cannot deceive grannies, that is, the situation when the ii-th granny in the moment of appearing in the courtyard, finds that now there are strictly less than aiai other grannies (except herself, but including Maria), is prohibited. Please note that if several grannies appeared in the yard at the same time, then each of them sees others at the time of appearance.
Your task is to find what maximum number of grannies (including herself) Maria can collect in the courtyard for the ceremony. After all, the more people in one place during quarantine, the more effective the ceremony!
Consider an example: if n=6n=6 and a=[1,5,4,5,1,9]a=[1,5,4,5,1,9], then:
- at the first step Maria can call grannies with numbers 11 and 55, each of them will see two grannies at the moment of going out into the yard (note that a1=1≤2a1=1≤2 and a5=1≤2a5=1≤2);
- at the second step, Maria can call grannies with numbers 22, 33 and 44, each of them will see five grannies at the moment of going out into the yard (note that a2=5≤5a2=5≤5, a3=4≤5a3=4≤5 and a4=5≤5a4=5≤5);
- the 66-th granny cannot be called into the yard — therefore, the answer is 66 (Maria herself and another 55 grannies).
Input
The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases in the input. Then test cases follow.
The first line of a test case contains a single integer nn (1≤n≤1051≤n≤105) — the number of grannies (Maria is not included in this number).
The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤2⋅1051≤ai≤2⋅105).
It is guaranteed that the sum of the values nn over all test cases of the input does not exceed 105105.Output
For each test case, print a single integer kk (1≤k≤n+11≤k≤n+1) — the maximum possible number of grannies in the courtyard.ExampleinputCopy
4 5 1 1 2 2 1 6 2 3 4 5 6 7 6 1 5 4 5 1 9 5 1 2 3 5 6
outputCopy
6 1 6 4
Note
In the first test case in the example, on the first step Maria can call all the grannies. Then each of them will see five grannies when they come out. Therefore, Maria and five other grannies will be in the yard.
In the second test case in the example, no one can be in the yard, so Maria will remain there alone.
The third test case in the example is described in the details above.
In the fourth test case in the example, on the first step Maria can call grannies with numbers 11, 22 and 33. If on the second step Maria calls 44 or 55 (one of them), then when a granny appears in the yard, she will see only four grannies (but it is forbidden). It means that Maria can’t call the 44-th granny or the 55-th granny separately (one of them). If she calls both: 44 and 55, then when they appear, they will see 4+1=54+1=5 grannies. Despite the fact that it is enough for the 44-th granny, the 55-th granny is not satisfied. So, Maria cannot call both the 44-th granny and the 55-th granny at the same time. That is, Maria and three grannies from the first step will be in the yard in total.
題意：每個人有一個聚會值，要求來了的人數（算同時來的）>=聚會值，問最多能拉到多少人
思路:和之前的cfdiv2一道分配人，每個人有經驗值很像，那道題是sort後從前往後尺取，這道題sort就從後往前貪心
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=2e5+10;
typedef long long LL;
LL a[maxn];
int main(void)
{
LL t;cin>>t;
while(t--)
{
LL n;cin>>n;
for(LL i=1;i<=n;i++) cin>>a[i];
sort(a+1,a+1+n);
if(a[n]<=n) cout<<n+1<<endl;
else
{
LL ans=1;LL sum=0;
for(LL i=n;i>=1;i--)
{
if(a[i]<=i)
{
sum+=i;break;
}
}
cout<<sum+1<<endl;
}
}
return 0;
}
C. Celex Update
During the quarantine, Sicromoft has more free time to create the new functions in “Celex-2021”. The developers made a new function GAZ-GIZ, which infinitely fills an infinite table to the right and down from the upper left corner as follows:
The cell with coordinates (x,y)(x,y) is at the intersection of xx-th row and yy-th column. Upper left cell (1,1)(1,1) contains an integer 11.
The developers of the SUM function don’t sleep either. Because of the boredom, they teamed up with the developers of the RAND function, so they added the ability to calculate the sum on an arbitrary path from one cell to another, moving down or right. Formally, from the cell (x,y)(x,y) in one step you can move to the cell (x+1,y)(x+1,y) or (x,y+1)(x,y+1).
After another Dinwows update, Levian started to study “Celex-2021” (because he wants to be an accountant!). After filling in the table with the GAZ-GIZ function, he asked you to calculate the quantity of possible different amounts on the path from a given cell (x1,y1)(x1,y1) to another given cell (x2,y2(x2,y2), if you can only move one cell down or right.
Formally, consider all the paths from the cell (x1,y1)(x1,y1) to cell (x2,y2)(x2,y2) such that each next cell in the path is located either to the down or to the right of the previous one. Calculate the number of different sums of elements for all such paths.Input
The first line contains one integer tt (1≤t≤571791≤t≤57179) — the number of test cases.
Each of the following tt lines contains four natural numbers x1x1, y1y1, x2x2, y2y2 (1≤x1≤x2≤1091≤x1≤x2≤109, 1≤y1≤y2≤1091≤y1≤y2≤109) — coordinates of the start and the end cells.Output
For each test case, in a separate line, print the number of possible different sums on the way from the start cell to the end cell.ExampleinputCopy
4 1 1 2 2 1 2 2 4 179 1 179 100000 5 7 5 7
outputCopy
2 3 1 1
Note
In the first test case there are two possible sums: 1+2+5=81+2+5=8 and 1+3+5=91+3+5=9.
題意：求(a,b)–>(c,d)的所有路徑總數中有多少路徑和不同的。
思路：當時自己是觀察規律的。賽後討論想出了證明
觀察發現，右上的數字比左下的小。那麼可以看到最大的一條路在要求的矩陣中是L形的，最小的一條路是對稱過去的反L 形的。
那麼我們要求的就轉化成了求 路的權值最大-路的權值最小的+1是多少。（中間的某個和是都能走出來的）
那麼對一個矩形來說，觀察最大和最小的和兩條路。
找最大路徑和最小路徑每一步的差,得出式子是 0(起點)+1+2+3+….x-1+x+x+x..+x-1+x-2+….3+2+1+0(終點);
對於1+2+3+..+x-1有兩個，求和得x*(x-1);
x有 (x+y+1)-x*2個。
最後得出 x*(y-x+1)+x*(x-1)==yx,所以一共有yx+1;
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5;
typedef long long LL;
int main(void)
{
LL t;cin>>t;
while(t--)
{
LL x1,y1,x2,y2;
cin>>x1>>y1>>x2>>y2;
cout<<(x2-x1)*(y2-y1)+1<<endl;
}
return 0;
}
D. The Best Vacation
You’ve been in love with Coronavirus-chan for a long time, but you didn’t know where she lived until now. And just now you found out that she lives in a faraway place called Naha.
You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly xx days and that’s the exact number of days you will spend visiting your friend. You will spend exactly xx consecutive (successive) days visiting Coronavirus-chan.
They use a very unusual calendar in Naha: there are nn months in a year, ii-th month lasts exactly didi days. Days in the ii-th month are numbered from 11 to didi. There are no leap years in Naha.
The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get jj hugs if you visit Coronavirus-chan on the jj-th day of the month.
You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).
Please note that your trip should not necessarily begin and end in the same year.Input
The first line of input contains two integers nn and xx (1≤n≤2⋅1051≤n≤2⋅105) — the number of months in the year and the number of days you can spend with your friend.
The second line contains nn integers d1,d2,…,dnd1,d2,…,dn, didi is the number of days in the ii-th month (1≤di≤1061≤di≤106).
It is guaranteed that 1≤x≤d1+d2+…+dn1≤x≤d1+d2+…+dn.Output
Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.ExamplesinputCopy
3 2 1 3 1
outputCopy
5
inputCopy
3 6 3 3 3
outputCopy
12
inputCopy
5 6 4 2 3 1 3
outputCopy
15
Note
In the first test case, the numbers of the days in a year are (indices of days in a corresponding month) {1,1,2,3,1}{1,1,2,3,1}. Coronavirus-chan will hug you the most if you come on the third day of the year: 2+3=52+3=5 hugs.
In the second test case, the numbers of the days are {1,2,3,1,2,3,1,2,3}{1,2,3,1,2,3,1,2,3}. You will get the most hugs if you arrive on the third day of the year: 3+1+2+3+1+2=123+1+2+3+1+2=12 hugs.
In the third test case, the numbers of the days are {1,2,3,4,1,2,1,2,3,1,1,2,3}{1,2,3,4,1,2,1,2,3,1,1,2,3}. You will get the most hugs if you come on the twelfth day of the year: your friend will hug you 2+3+1+2+3+4=152+3+1+2+3+4=15 times.
題意：每個月有d[i]天，有x天度假，要求連續，可以跨年，保證度假日子小於一整年。求d[i]求和最大的一段
思路：貪心+前綴和+二分
貪心：
有這麼一段區間，貪A和貪B。
當bn(月的最後一天)>an-2,明顯可以得出B是最優的。
那麼當bn<an-2時，可以推出an-2>bn—->(等差遞增)an>bn,而且a月比b月長且最大值更大，所以在a月的末尾作爲最後一天最優。
所以貪心把末尾的一天放到月末。
然後我們利用前綴和統計每個月的天數和價值。然後枚舉右端點，二分左端點，找到
sum[i]-sum[l]>x;且sum[i]-sum[l+1]<=x
那麼價值爲summoney[i]-summoney[l+1];再算多出的x部分–知道x多出有多少天，然後在多出的區間裏用等差數列公式算出來。比如多出3天，就是那段區間的後三天的價值和。
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=4e5+20;
typedef long long LL;
LL a[maxn];
LL sumday[maxn];
LL summoney[maxn];
int main(void)
{
LL n,x;cin>>n>>x;
for(LL i=1;i<=n;i++)
{
cin>>a[i];
sumday[i]=sumday[i-1]+a[i];
summoney[i]=summoney[i-1]+(1+a[i])*a[i]/2;
}
for(LL i=n+1;i<=2*n;i++)
{
a[i]=a[i-n];
sumday[i]=sumday[i-1]+a[i];
summoney[i]=summoney[i-1]+(1+a[i])*a[i]/2;
}
LL res=0;
for(LL i=1;i<=2*n;i++)
{
LL l=0;LL r=i+1;
//邊界處理
//找出第一個區間差大於等於x的位置
while(l<r)
{
LL mid=(l+r)>>1;
if(sumday[i]-sumday[mid]<=x) r=mid;
else l=mid+1;
}
if(l==0||r==i+1) continue;
LL resday=x-(sumday[i]-sumday[l]);
LL sum=summoney[i]-summoney[l];
if(a[l]>=resday) sum+=(2*a[l]-resday+1)*resday/2;
res=max(res,sum);
}
cout<<res<<endl;
return 0;
}