title: “第一场补题”
date: “2020-01-09 17:17”

###D - Eat Candies
You have three piles of candies: red, green and blue candies:

the first pile contains only red candies and there are r candies in it,
the second pile contains only green candies and there are g candies in it,
the third pile contains only blue candies and there are b 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 t (1≤t≤1000) — the number of test cases in the input. Then t test cases follow.

Each test case is given as a separate line of the input. It contains three integers r, g and b (1≤r,g,b≤108) — the number of red, green and blue candies, respectively.

Output
Print t integers: the i-th printed integer is the answer on the i-th test case in the input.

Example
Input
6
1 1 1
1 2 1
4 1 1
7 4 10
8 1 4
8 2 8
Output
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.

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main(){int n;int a[3];cin>>n;while(n--){cin>>a[0]>>a[1]>>a[2];sort(a,a+3);if(a[2]>a[0]+a[1])cout<<a[0]+a[1]<<endl;elsecout<<(a[0]+a[1]+a[2])/2<<endl;}
return 0;
}

###E - 由你来决定怎么颁奖

So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved pi problems. Since the participants are primarily sorted by the number of solved problems, then p1≥p2≥⋯≥pn.

Help the jury distribute the gold, silver and bronze medals. Let their numbers be g, s and b, respectively. Here is a list of requirements from the rules, which all must be satisfied:

for each of the three types of medals, at least one medal must be awarded (that is, g>0, s>0 and b>0);
the number of gold medals must be strictly less than the number of silver and the number of bronze (that is, g<s and g<b, but there are no requirements between s and b);
each gold medalist must solve strictly more problems than any awarded with a silver medal;
each silver medalist must solve strictly more problems than any awarded a bronze medal;
each bronze medalist must solve strictly more problems than any participant not awarded a medal;
the total number of medalists g+s+b should not exceed half of all participants (for example, if n=21, then you can award a maximum of 10 participants, and if n=26, then you can award a maximum of 13 participants).
The jury wants to reward with medals the total maximal number participants (i.e. to maximize g+s+b) so that all of the items listed above are fulfilled. Help the jury find such a way to award medals.

Input
The first line of the input contains an integer t (1≤t≤10000) — the number of test cases in the input. Then t test cases follow.

The first line of a test case contains an integer n (1≤n≤4⋅105) — the number of BeRC participants. The second line of a test case contains integers p1,p2,…,pn (0≤pi≤106), where pi is equal to the number of problems solved by the i-th participant from the final standings. The values pi are sorted in non-increasing order, i.e. p1≥p2≥⋯≥pn.

The sum of n over all test cases in the input does not exceed 4⋅105.

Output
Print t lines, the j-th line should contain the answer to the j-th test case.

The answer consists of three non-negative integers g,s,b.

Print g=s=b=0 if there is no way to reward participants with medals so that all requirements from the statement are satisfied at the same time.
Otherwise, print three positive numbers g,s,b — the possible number of gold, silver and bronze medals, respectively. The sum of g+s+b should be the maximum possible. If there are several answers, print any of them.
Example
Input
5
12
5 4 4 3 2 2 1 1 1 1 1 1
4
4 3 2 1
1
1000000
20
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
32
64 64 63 58 58 58 58 58 37 37 37 37 34 34 28 28 28 28 28 28 24 24 19 17 17 17 17 16 16 16 16 11
Output
1 2 3
0 0 0
0 0 0
2 5 3
2 6 6
Note
In the first test case, it is possible to reward 1 gold, 2 silver and 3 bronze medals. In this case, the participant solved 5 tasks will be rewarded with the gold medal, participants solved 4 tasks will be rewarded with silver medals, participants solved 2 or 3 tasks will be rewarded with bronze medals. Participants solved exactly 1 task won’t be rewarded. It’s easy to see, that in this case, all conditions are satisfied and it is possible to reward participants in this way. It is impossible to give more than 6 medals because the number of medals should not exceed half of the number of participants. The answer 1, 3, 2 is also correct in this test case.

In the second and third test cases, it is impossible to reward medals, because at least one medal of each type should be given, but the number of medals should not exceed half of the number of participants.

#include<iostream>
#include<cstdio>
using namespace std;
int main(){int n;cin>>n;while(n--){int t;int g=0,s=0,b=0;int a[400010];cin>>t;int sum=t/2;for(int i = 1; i <= t; i++){cin>>a[i];}int goal=a[sum+1];for(int i=1;i <= sum ;i++){if(a[i]==goal){sum=i-1;break;}}// for(int i=1;i<=sum;i++){//     cout<<a[i]<<" ";// }// cout<<endl;int flag=1;while( a[flag] == a[1] && flag <= sum ){g++;flag++;}int tt=a[flag];while(tt==a[flag]&&flag<=sum){s++;flag++;}while(s<=g&&flag<=sum){tt=a[flag];while(a[flag]==tt && flag<=sum){s++;flag++;}}b = sum - g -s; //sum+1可能和sum一样// while(a[flag]!=a[sum+1]&&flag<=sum){//     b++;//     flag++;// }//  if(g>=s||g>=b){// printf("0 0 0\n");// 	continue;// }// printf("%d %d %d\n",g,s,b);if(g > 0 && s > 0 && b > 0 && g < s && g < b){cout<<g<<" "<<s<<" "<<b<<endl;}elsecout<<0<<" "<<0<<" "<<0<<endl;}return  0;}

###F - XorXor
Alex has a sequence S which contains n elements, let’s define f(i, j) denoting the xor sum of Si, Si+1, …, Sj. Now Alex wants to know the value of f(1, 1) xor f(1, 2) xor … xor f(1, n) xor f(2, 2) xor f(2, 3) xor … xor f(2, n) xor … xor f(n, n).
Input
Input starts with an integer T denoting the number of test case.
For each test case, first line contains an integer n (1 <= n <= 100000).
Next line contains n integers Si(1 <= Si <= 100000).
Output
For each test case. print the value of
f(1, 1) xor f(1, 2) xor … xor f(1, n) xor f(2, 2) xor f(2, 3) xor … xor f(2, n) xor … xor f(n, n).
Sample Input
1
3
1 3 2
Sample Output
3
Hint
无

#include<iostream>
#include<cstdio>
using namespace std;
int main(){int n;scanf("%d",&n);while(n--){int t;int a;int res=0;scanf("%d",&t);for(int i=1;i<=t;i++){scanf("%d",&a);if(i*(t-i+1)&1){res^=a;}}printf("%d\n",res);}return 0;
}

###G - 0011
Alex likes to play with one and zero!One day he gets an empty string.So our cute boy wants to add one and zero in it. Every time he will add ‘01’in the string at any position and then get a new string.For example:if the string is “01” now ,he can get “0101” or “0011,Now give you a string that is Alex has get,you need to answer whether the string is legal?

Input
First is a integer n(n<=100)
Next contains n lines .Every line is a string whose legth is no more than 1000.
Output
For each case output “YES” in a single line if it’s legal.
Or you need to output “NO”;
Sample Input
3
0101
0110
0011
Sample Output
YES
NO
YES
Hint
无

#include<iostream>
#include<stack>
#include<cstring>
#include<cstdio>
using namespace std;
int main(){int n;cin>>n;while(n--){   string tt;//char tt[1010];stack<char> ss;cin>>tt;//scanf("%s",tt);//int len=strlen(tt);int len=tt.length();for(int gg=0;gg<len;gg++){if(ss.empty()){ss.push(tt[gg]);}else if(ss.top()=='0'&&tt[gg]=='1'){ss.pop();}else{ss.push(tt[gg]);}}if(ss.empty())cout<<"YES"<<endl;elsecout<<"NO"<<endl;}
return 0;
}

###H - Perfect String
A string is called beautiful if no two consecutive characters are equal. For example, “ababcb”, “a” and “abab” are beautiful strings, while “aaaaaa”, “abaa” and “bb” are not.

Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters ‘a’, ‘b’, ‘c’ and ‘?’. Ahcl needs to replace each character ‘?’ with one of the three characters ‘a’, ‘b’ or ‘c’, such that the resulting string is beautiful. Please help him!

More formally, after replacing all characters ‘?’, the condition si≠si+1 should be satisfied for all 1≤i≤|s|−1, where |s| is the length of the string s.

Input
The first line contains positive integer t (1≤t≤1000) — the number of test cases. Next t lines contain the descriptions of test cases.

Each line contains a non-empty string s consisting of only characters ‘a’, ‘b’, ‘c’ and ‘?’.

It is guaranteed that in each test case a string s has at least one character ‘?’. The sum of lengths of strings s in all test cases does not exceed 105.

Output
For each test case given in the input print the answer in the following format:

If it is impossible to create a beautiful string, print “-1” (without quotes);
Otherwise, print the resulting beautiful string after replacing all ‘?’ characters. If there are multiple answers, you can print any of them.
Example
Input
3
a???cb
a??bbc
a?b?c
Output
ababcb
-1
acbac
Note
In the first test case, all possible correct answers are “ababcb”, “abcacb”, “abcbcb”, “acabcb” and “acbacb”. The two answers “abcbab” and “abaabc” are incorrect, because you can replace only ‘?’ characters and the resulting string must be beautiful.

In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal.

In the third test case, the only answer is “acbac”.

#include<iostream>
#include<cstring>
using namespace std;
int main(){int n;cin>>n;string t;while(n--){int flag=0;cin>>t;if(t[0]=='?'){if(t[1]!='a') t[0]='a';else if(t[1]!='b') t[0]='b';else if(t[1]!='c') t[0]='c';}for(int i=0;i<t.length();i++){if(t[i]=='?'){if(t[i+1]!='a'&&t[i-1]!='a') t[i]='a';else if(t[i+1]!='b'&&t[i-1]!='b') t[i]='b';else if(t[i+1]!='c'&&t[i-1]!='c') t[i]='c';}else if(t[i]==t[i+1]){flag=1;break;}}if(flag) cout<<-1<<endl;else {cout<<t<<endl;}}return 0;
}

###I - 十进制中的二进制
hkhv学长最近对二进制数很感兴趣，喜欢一切0和1组成的数。现在有一个十进制整数n，问你1到n之间有多少个数是只有0和1组成的类似二进制的数，输出他们的个数。

Input
输入数据包含一个数n (1 <= n <=10^9).
Output
输出1到n中类似二进制的数的个数.
Sample Input
10
Sample Output
2
Hint
对于n = 10，1 和 10是类似二进制的数.

#include<iostream>
#include<cstdio>
using namespace std;
long long int n;
int a[2]={0,1};
int sum=0;
void dfs(long long int x){if(x>n) return ;else{sum++;for(int i=0;i<=1;i++){dfs(x*10+a[i]);}}
}
int main(){while(cin>>n){sum=0;dfs(1);cout<<sum<<endl;}
return 0;
}

###Frog Wa has many frogs, so he can hear GuaGuaGua every day. But one day a flappy bird ate his frogs, so Frog Wa decided to build a fence to protect his precious frogs. Frog Wa told you four points (x, 0), (x, y), (z, 0), (z, t) which represent for the vertex of the fence. he want to know the area of the fence, can you tell him?
Input
The input consist of four integers x, y, z and t. 0 <= x, y, z, t <= 10000, x < z and y < t.
Output
For each test case, print the area, numbers should accurate to one decimal places.
Sample Input
1 1 2 2
0 2 6 6
0 0 2 10
Sample Output
1.5
24.0
10.0
Hint
无

#include<iostream>
#include<cstdio>
#define long long ll
using namespace std;
int main(){double x,y,z,t;while(cin>>x>>y>>z>>t){double s=(y+t)*(z-x)/2;printf("%.1lf\n",s);}return 0;
}