B - Everyone is a Winner!
On the well-known testing system MathForces, a draw of n rating units is arranged. The rating will be distributed according to the following algorithm: if k participants take part in this event, then the n rating is evenly distributed between them and rounded to the nearest lower integer, At the end of the drawing, an unused rating may remain — it is not given to any of the participants.
For example, if n=5 and k=3, then each participant will recieve an 1 rating unit, and also 2 rating units will remain unused. If n=5, and k=6, then none of the participants will increase their rating.
Vasya participates in this rating draw but does not have information on the total number of participants in this event. Therefore, he wants to know what different values of the rating increment are possible to get as a result of this draw and asks you for help.
For example, if n=5, then the answer is equal to the sequence 0,1,2,5. Each of the sequence values (and only them) can be obtained as ⌊n/k⌋ for some positive integer k (where ⌊x⌋ is the value of x rounded down): 0=⌊5/7⌋, 1=⌊5/5⌋, 2=⌊5/2⌋, 5=⌊5/1⌋.
Write a program that, for a given n, finds a sequence of all possible rating increments.
Input
The first line contains integer number t (1≤t≤10) — the number of test cases in the input. Then t test cases follow.
Each line contains an integer n (1≤n≤109) — the total number of the rating units being drawn.
Output
Output the answers for each of t test cases. Each answer should be contained in two lines.
In the first line print a single integer m — the number of different rating increment values that Vasya can get.
In the following line print m integers in ascending order — the values of possible rating increments.
Sample Input
4
5
11
1
3
Sample Output
4
0 1 2 5
6
0 1 2 3 5 11
2
0 1
3
0 1 3
一共有n块钱，k个人[n/k]块钱，向下取整。
现在给你n块钱，你不知道有多少人，输出每个人可能获得多少钱
set集合中没有重复的元素
set容器内所有元素都是以节点的方式来存储，其节点结构和链表差不多，指向父节点和子节点
因此插入的时候只需稍作变换，把节点的指针指向新的节点就可以了，删除类似
这里的一切操作都是指针的变换，和内存移动是没有关系的
在set中查找是使用二分查找
#include<iostream>
#include<stdio.h>
#include<set>
using namespace std;
int main(){
int t;
cin>>t;
for(int i=1;i<=t;i++){
int n;
cin>>n;
set<int> s;
for(int i=1;i*i<=n;i++){
s.insert(i);
s.insert(n/i);
}
cout<<s.size()+1<<endl;
s.insert(0);
for(auto x:s){
cout<<x<<" ";
}
cout<<endl;
}
}
E - PIN Codes
outputstandard output
A PIN code is a string that consists of exactly 44 digits. Examples of possible PIN codes: 7013, 0000 and 0990. Please note that the PIN code can begin with any digit, even with 0.
Polycarp has nn (2≤n≤102≤n≤10) bank cards, the PIN code of the ii-th card is pipi.
Polycarp has recently read a recommendation that it is better to set different PIN codes on different cards. Thus he wants to change the minimal number of digits in the PIN codes of his cards so that all nn codes would become different.
Formally, in one step, Polycarp picks ii-th card (1≤i≤n1≤i≤n), then in its PIN code pipi selects one position (from 11 to 44), and changes the digit in this position to any other. He needs to change the minimum number of digits so that all PIN codes become different.
Polycarp quickly solved this problem. Can you solve it?
Input
The first line contains integer tt (1≤t≤1001≤t≤100) — the number of test cases in the input. Then test cases follow.
The first line of each of tt test sets contains a single integer nn (2≤n≤102≤n≤10) — the number of Polycarp's bank cards. The next nn lines contain the PIN codes p1,p2,…,pnp1,p2,…,pn — one per line. The length of each of them is 44. All PIN codes consist of digits only.
Output
Print the answers to tt test sets. The answer to each set should consist of a n+1n+1 lines
In the first line print kk — the least number of changes to make all PIN codes different. In the next nn lines output the changed PIN codes in the order corresponding to their appearance in the input. If there are several optimal answers, print any of them.
Example
input
Copy
3
2
1234
0600
2
1337
1337
4
3139
3139
3139
3139
output
Copy
0
1234
0600
1
1337
1237
3
3139
3138
3939
6139
第一个可以称为关键字，每个关键字只能在map中出现一次
第二个可能称为该关键字的值
map以模板方式实现，可以存储任意类型的数据，包括使用者自定义的数据类型。Map主要用于
一对一映射的情况，map内部的实现自建一颗红黑树，这棵树具有对数据自动排序的功能，在map4
内部所有的数据都是有序，后面我们会见识到有序的好处，
map对象是模板类，需要关键字和存储对象两个模板参数
#include<iostream>
#include<stdio.h>
#include<map>
using namespace std;
int h[15];
int main(){
int t;
cin>>t;
for(int i=1;i<=t;i++){
int ans = 0;
int n;
cin>>n;
map<int,int>A;
map<int,int>B;
for(int i=1;i<=n;i++){
scanf("%d",&h[i]);
A[h[i]]++;
B[h[i]%10]++;
if(A[h[i]]>=2){
ans++;
}
}
cout<<ans<<endl;
for(int i = 1;i<=n;i++){
if(A[h[i]]>=2){
A[h[i]]--;
int g = h[i];
int m1 = g%10; g = g/10;
int m2 = g%10; g = g/10;
int m3 = g%10; g=g/10;
int m4 = g;
for(int j = 0;j<=9;j++){
if(B[j]==0){
B[j]++;
cout<<m4<<m3<<m2<<j<<endl;
break;
}
}
}
else{
printf("%04d
",h[i]);//保证当第一位为0的情况
}
}
}
}
C - Sweet Problem
You have three piles of candies: red, green and blue candies:
the first pile contains only red candies and there are rr candies in it,
the second pile contains only green candies and there are gg candies in it,
the third pile contains only blue candies and there are bb candies in it.
Each day Tanya eats exactly two candies of different colors. She is free to choose the colors of eaten candies: the only restriction that she can’t eat two candies of the same color in a day.
Find the maximal number of days Tanya can eat candies? Each day she needs to eat exactly two candies.
Input
The first line contains integer tt (1≤t≤10001≤t≤1000) — the number of test cases in the input. Then tt test cases follow.
Each test case is given as a separate line of the input. It contains three integers rr, gg and bb (1≤r,g,b≤1081≤r,g,b≤108) — the number of red, green and blue candies, respectively.
Output
Print tt integers: the ii-th printed integer is the answer on the ii-th test case in the input.
Example
input
Copy
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
output
Copy
1
2
2
10
5
9
Note
In the first example, Tanya can eat candies for one day only. She can eat any pair of candies this day because all of them have different colors.
In the second example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and green and blue candies on the second day.
In the third example, Tanya can eat candies for two days. For example, she can eat red and green candies on the first day, and red and blue candies on the second day. Note, that two red candies will remain uneaten.
题意：
有 r 个红糖果，g 个绿糖果，b 个蓝糖果，1 <= r,g,b <= 10^8. 每天只能吃两个糖果，问最多可以吃多少天。
思路：
先找出数量最多的哪一种糖果，如果那种糖果的数量大于另外两种糖果的数量之和，则答案必然是另外两种的糖果的数量和，因为即使是每天都吃数量最多的糖果，最后也必然会只剩下哪一种糖果。
如果数量最多的糖果的数量小于等于另外两种糖果数量之和，则输出三者的数量和除以2的结果（小数则向下取整）。因为对于任意满足条件的三个数a,b,c（降序），可以先使a与b配对，剩下的与c配对，最后必然会有c剩下，然后不断拆掉a,b的一对，至c<=1;若a+b+c为奇数，则必然剩下一颗，为偶数则必然全部吃完。
代码：
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
int main(){
ll t,a[3];
scanf("%lld",&t);
while(t--){
scanf("%lld%lld%lld",&a[0],&a[1],&a[2]);
sort(a,a+3);
if(a[2]>a[0]+a[1])
printf("%lld
",a[1]+a[0]);
else
printf("%lld
",(a[0]+a[1]+a[2])/2);
}
return 0;
}
A - Secret Passwords
One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of nn passwords — strings, consists of small Latin letters.
Hacker went home and started preparing to hack AtForces. He found that the system contains only passwords from the stolen list and that the system determines the equivalence of the passwords aa and bb as follows:
two passwords aa and bb are equivalent if there is a letter, that exists in both aa and bb;
two passwords aa and bb are equivalent if there is a password cc from the list, which is equivalent to both aa and bb.
If a password is set in the system and an equivalent one is applied to access the system, then the user is accessed into the system.
For example, if the list contain passwords "a", "b", "ab", "d", then passwords "a", "b", "ab" are equivalent to each other, but the password "d" is not equivalent to any other password from list. In other words, if:
admin's password is "b", then you can access to system by using any of this passwords: "a", "b", "ab";
admin's password is "d", then you can access to system by using only "d".
Only one password from the list is the admin's password from the testing system. Help hacker to calculate the minimal number of passwords, required to guaranteed access to the system. Keep in mind that the hacker does not know which password is set in the system.
Input
The first line contain integer nn (1≤n≤2⋅1051≤n≤2⋅105) — number of passwords in the list. Next nn lines contains passwords from the list – non-empty strings sisi, with length at most 5050 letters. Some of the passwords may be equal.
It is guaranteed that the total length of all passwords does not exceed 106106 letters. All of them consist only of lowercase Latin letters.
Output
In a single line print the minimal number of passwords, the use of which will allow guaranteed to access the system.
Examples
input
Copy
4
a
b
ab
d
output
Copy
2
input
Copy
3
ab
bc
abc
output
Copy
1
input
Copy
1
codeforces
output
Copy
1
思路：
并查集
代码：
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int maxn = 51;
char a[maxn];
int f[maxn];
bool vis[maxn];
int ans=0,n;
inline int find(int x){
return f[x]==x?x:f[x]=find(f[x]);
}
void merge(int x,int y){
x = find(x);
y = find(y);
if(x!=y) f[x] = y;
//return ;
}
int main(){
int n;
cin>>n;
for(int i=1;i<=26;i++){
f[i] = i;
}
for(int i = 1;i<=n;i++){
scanf("%s",a+1);
vis[a[1]-'a'+1] = true;
int len = strlen(a+1);
for(int j=2;j<=len;j++){
merge(a[j-1]-'a'+1,a[j]-'a'+1);
vis[a[j]-'a'+1] = true;
}
}
for(int i = 1;i<=26;i++){
if(vis[i]&&(f[i]==i))
ans++;
}
cout<<ans<<endl;
}