7-2 Broken Pad (20 分)

1.题目描述：

The party began, the greasy uncle was playing cards, the fat otaku was eating, and the little beauty was drawing.

Playing cards is an indispensable and irreplaceable activity for parties. In order to be more impartial and prevent magician YHH from cheating when shuffling cards, and give full play to his mind-reading advantages at the same time, psychologist ZH proposed to use a pad to play cards.

However, ZH found that the touch screen of the pad he was assigned was malfunctioning. Every time he clicked a card, this card and all the cards behind it would be selected. As we all know, the effect of choosing a card is equivalent to changing the state of the card, that is, if the card was initially selected, it would become unselected, and if it was unselected, it would become selected. Besides, there is another operation, which is to click on the blank space, so that all the cards will become unselected.

Now ZH needs to select some cards to play, but due to the malfunctioning of the touch screen, he cannot simply choose the cards he wants to choose. Now he has used the blind trick to secretly ask netizens which positions to click to choose the cards he wants, please help him!

Given two 01-strings a and b, which represent the current state of the card and the state required by ZH. 0 represents unselected, and 1 represents selected. Now you are asked to give a plan with the smallest number of tap times.

Input Specification:

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 10), indicating the number of test cases. For each test case:

The first line contains a string a (1 ≤ ∣a∣ ≤ 105), indicating the current state of the card.

The second line contains a string b (∣b∣=∣a∣), indicating the state required by ZH.

Output Specification:

For each test case, print one line contains the minimum number of integers indicating the position ZH should tap in non-decreasing order.

Please note that number 0 is indicating the blank space, and it's guaranteed that the solution of all the test cases is unique.

Sample Input:

2
10110
10000
110101
000000

结尾无空行

Sample Output:

3 5
0

结尾无空行

Hint:

For the first sample, a is "10110", and b is "10000", then ZH needs first to tap the position 3 (based on 1) to make the state become "10001", and then tap the position 5 to make the state become "10000", so you should tell him 3 and 5.

  2.代码展示：

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e5+5;
//用于计算操作数 
int cal(string s1,string s2,int c[]){
	int count1=0,num=0;
    int flag=1;//用于标记状态是否改变 
	for(int i=0;s1[i];i++){
		    //如果此时片初始状态和终止状态不同，且卡片初始状态未改变 
			if((s1[i]!=s2[i])&&flag){
                c[num++]=i+1;//存储操作过程 
				count1++;//统计操作数 
                flag=0;//改变状态 
			}
			//如果此时片初始状态和终止状态相同，且卡片初始状态改变
			else if(!flag&&(s1[i]==s2[i])){
                c[num++]=i+1;//存储操作过程 
				count1++;//统计操作数 
                flag=1;//改变状态 
            }
	}
	return count1;
}
int main(){
	int T;//T为测试用例的数量
	cin>>T;
	while(T--){
		string s1,s2,s3;
        int a[MAX],b[MAX];
		cin>>s1>>s2;
		s3=s1;
		int count1=0,count2=0;
		//1.按清零操作 
		for(int i=0;s1[i];i++)s1[i]='0';//将卡片初始状态全部清零 
        count1++;//操作数加1 
		count1=cal(s1,s2,a);//进行操作数的计算 
		//2.不按清零操作 
		count2=cal(s3,s2,b);//进行操作数的计算 //进行操作数的计算 
		//比较清零与不清零的操作数，看看谁是最小值 
		if(count1<count2){
			cout<<"0";
			for(int i=0;i<count1;i++)cout<<" "<<a[i];
		}else {
            for(int i=0;i<count2;i++){
                if(i==count2-1)cout<<b[i];
                else cout<<b[i]<<" ";
            }
        }
		cout<<endl;	
	}
	return 0;
}

原题链接：PTA | 程序设计类实验辅助教学平台