CONTINUE...?


Time Limit: 1 Second      Memory Limit: 65536 KB      Special Judge

DreamGrid has  classmates numbered from  to . Some of them are boys and the others are girls. Each classmate has some gems, and more specifically, the -th classmate has  gems.

DreamGrid would like to divide the classmates into four groups , ,  and  such that:

  • Each classmate belongs to exactly one group.

  • Both  and  consist only of girls. Both  and  consist only of boys.

  • The total number of gems in  and  is equal to the total number of gems in  and .

Your task is to help DreamGrid group his classmates so that the above conditions are satisfied. Note that you are allowed to leave some groups empty.

Input

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

The first line contains an integer  () -- the number of classmates.

The second line contains a string  () consisting of 0 and 1. Let  be the -th character in the string . If , the -th classmate is a boy; If , the -th classmate is a girl.

It is guaranteed that the sum of all  does not exceed .

Output

For each test case, output a string consists only of {1, 2, 3, 4}. The -th character in the string denotes the group which the -th classmate belongs to. If there are multiple valid answers, you can print any of them; If there is no valid answer, output "-1" (without quotes) instead.

Sample Input

5
1
1
2
10
3
101
4
0000
7
1101001

Sample Output

-1
-1
314
1221
3413214 原题地址:http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5759
题意:
老师分宝石,一个由0和1组成的字符串数组,每个字符都是一个学生每个学生s[i]都有i个宝石,其中1是男生,0是女生
有G1G2G3G4四个分组其中G1G2只能女生,G2G3只能男生。要求G1G3一组,G2G4一组,让他们两组的宝石个数相同 思路:
一个数组能否被分成两堆第一步是取决于他们的和是否为偶数,如果是奇数是不可能分成两堆的。
题目有说可以让有的组数为0所以男生女生这个要求就只是一个干扰项;
具体:首先我们判断长度为偶数的情况
例如:
8
11111111
这个数据
首先他们的值是
1 2 3 4 5 6 7 8
我们让第一个”1“为第A组,第二个”2“为第B组 这样第A组的值减去第B组的值为(1-2)=-1;
然后我们让第三个”3“为第B组,第四个”4“为第A组,这样第A组减去第B组的值为(4-3)=1;
正好抵消为0就这样依次 A B B A A B B A就能保证他们的合为0; 然后我们看奇数长度的情况
例如:题目的最后一组数据
7
1101001
他们的值是
1 2 3 4 5 6 7
首先我们从第二个开始
第二个在A组,第三个在B组 这时候A-B=(2-3)=-1;
第四个在B组,第五个在A组 这时候A-B=(5-4)=1; 这时前面四个正好抵消
然后第六个在A组,第四个在B组,这时候A-B=(6-7)=-1;
这时候正好可以把前面被忽略的1加进来放在不够的A组上这时候正好抵消; 代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int one;
int two;
int three;
int four;
int mark[];
bool yes(ll n){
if(((n*(n+))/)%==)return false;
return true;
}
int main()
{
//std::ios::sync_with_stdio(false);
ll n;
int t;
cin>>t;
string s;
while(t--){
cin>>n;
cin>>s;
if(!yes(n)){//判断是否合为偶数
cout<<-<<endl;
continue;
}
if(n%==){///长度为偶数的情况
for(int i=;i<n;i++){
if(i%==||i%==){
if(s[i]=='')cout<<"";
else cout<<"";
}
else if(i%==||i%==){
if(s[i]=='')cout<<"";
else cout<<"";
}
}
}
else{///长度为奇数的情况
if(s[]=='')cout<<"";
else cout<<"";
for(int i=;i<n;i++){
if((i-)%==||(i-)%==){
if(s[i]=='')cout<<"";
else cout<<"";
}
else if((i-)%==||(i-)%==){
if(s[i]=='')cout<<"";
else cout<<"";
}
}
}
cout<<endl;
}
return ;
}

The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple - J CONTINUE...?的更多相关文章

  1. The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple - L Doki Doki Literature Club

    Doki Doki Literature Club Time Limit: 1 Second      Memory Limit: 65536 KB Doki Doki Literature Club ...

  2. 2018浙江省赛(ACM) The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple

    我是铁牌选手 这次比赛非常得爆炸,可以说体验极差,是这辈子自己最脑残的事情之一. 天时,地利,人和一样没有,而且自己早早地就想好了甩锅的套路. 按理说不开K就不会这么惨了啊,而且自己也是毒,不知道段错 ...

  3. The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple - M Lucky 7

    Lucky 7 Time Limit: 1 Second      Memory Limit: 65536 KB BaoBao has just found a positive integer se ...

  4. The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple - B King of Karaoke

    King of Karaoke Time Limit: 1 Second      Memory Limit: 65536 KB It's Karaoke time! DreamGrid is per ...

  5. The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple -A Peak

    Peak Time Limit: 1 Second      Memory Limit: 65536 KB A sequence of  integers  is called a peak, if ...

  6. ZOJ 4033 CONTINUE...?(The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple)

    #include <iostream> #include <algorithm> using namespace std; ; int a[maxn]; int main(){ ...

  7. The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple - F 贪心+二分

    Heap Partition Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge A sequence S = { ...

  8. The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple - C 暴力 STL

    What Kind of Friends Are You? Time Limit: 1 Second      Memory Limit: 65536 KB Japari Park is a larg ...

  9. ZOJ 3962 E.Seven Segment Display / The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple E.数位dp

    Seven Segment Display Time Limit: 1 Second      Memory Limit: 65536 KB A seven segment display, or s ...

随机推荐

  1. MFC Button控件自绘

    文章参考地址:  http://blog.csdn.net/yue7603835/article/details/6649458    VC下的界面着实难看 有时候我们不得不自己进行控件的绘制 以前 ...

  2. Maven如何打包本地依赖包

    有的jar包,在maven中心库里面是没有的,那么,如何在项目中使用呢? 假设我们需要使用:apache-ant-zip-2.3.jar 将该jar包,放在项目的lib目录,例如: 在pom.xml里 ...

  3. istringstream()函数的用法

    istringstream()函数的用法 头文件:#include 功能:将一个含有多个空格的字符串分割开来 eg:

  4. 转载:Apache commons开源工具简介

    Apache Commons是一个非常有用的工具包,解决各种实际的通用问题,下面是一个简述表,详细信息访问http://jakarta.apache.org/commons/index.html Be ...

  5. 【BZOJ2663】灵魂宝石 [二分]

    灵魂宝石 Time Limit: 5 Sec  Memory Limit: 128 MB[Submit][Status][Discuss] Description “作为你们本体的灵魂,为了能够更好的 ...

  6. bzoj 1798 维护序列seq 线段树

    裸的线段树,注意标签下放就行了 多么痛的领悟,一定要开int64 /************************************************************** Pro ...

  7. SELinux 案例 1

    deamon 进程的权限白名单 Generate Policy for Internet Services Daemon,生成一个模板: sepolicy generate --inetd [-n N ...

  8. LARGE_INTEGER类型

    最近在研究Windows驱动程序开发,遇到这样一个语句: devExt->PortBase  = (PUCHAR)(ULONG_PTR) portBasePA.QuadPart; 在源代码环境中 ...

  9. [Leetcode Week5]Word Ladder II

    Word Ladder II 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/word-ladder-ii/description/ Descripti ...

  10. JSTL c:url

    c:url 标签 jstl 实例代码和用法.     <c:url>标记格式化成一个字符串格式的URL,并将其存储到变量中.这个标签会在必要时自动执行URL重写. var属性指定的变量将包 ...