Codeforces Round 920 (Div. 3) B. Arranging Cats

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#c++ #开发语言

Problem Statement

In order to test the hypothesis about the cats, the scientists must arrange the cats in the boxes in a specific way. Of course, they would like to test the hypothesis and publish a sensational article as quickly as possible, because they are too engrossed in the next hypothesis about the phone’s battery charge.

Scientists have n boxes in which cats may or may not sit. Let the current state of the boxes be denoted by the sequence b1,…,bn: bi=1 if there is a cat in box number i, and bi=0 otherwise.

Fortunately, the unlimited production of cats has already been established, so in one day, the scientists can perform one of the following operations:

Take a new cat and place it in a box (for some i such that bi=0, assign bi=1).
Remove a cat from a box and send it into retirement (for some i such that bi=1, assign bi=0).
Move a cat from one box to another (for some i,j such that bi=1,bj=0, assign bi=0,bj=1).

It has also been found that some boxes were immediately filled with cats. Therefore, the scientists know the initial position of the cats in the boxes s1,…,sn and the desired position f1,…,fn

Due to the large amount of paperwork, the scientists do not have time to solve this problem. Help them for the sake of science and indicate the minimum number of days required to test the hypothesis.

Input

Each test consists of several test cases. The first line contains a single integer t (1≤t≤104) — the number of test cases. This is followed by descriptions of the test cases.

Each test case consists of three lines.

The first line of each test case contains a single integer n (1≤n≤105) — the number of boxes.

The second line of each test case contains a strings
of n characters, where the i-th character is ‘1’ if there is a cat in the i-th box and ‘0’ otherwise.

The third line of each test case contains a string f
of n characters, where the i-th character is ‘1’ if there should be a cat in the i-th box and ‘0’ otherwise.

It is guaranteed that in a test the sum of n over all test cases does not exceed 105.

Output

For each test case, output a single integer on a separate line — the minimum number of operations required to obtain the desired position from the initial position. It can be shown that a solution always exists.

SAMPLES

input

6 5 10010 00001 1 1 1 3 000 111 4 0101 1010 3 100 101 8 10011001 11111110

output

2 0 3 2 1 4

解题过程

思路：先记录第一次输入数1的位置，然后再与第二次输入作比较，两次输入相同位置一样的位置可以不用考虑

下面是一种方法，用一个数组vis[]来记录第一次输入0和1的状态，第二次仅比较

代码段

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        int t;
        cin>>t;
        int x;
        int vis[N];
        int n1=0,n2=0;
        for(int i=1;i<=t;i++)
        {
            scanf("%1d",&x);
            if(x==1)
            {
                vis[i]=1;
                n1++;
            }
            else
                vis[i]=0;
        }
        for(int i=1;i<=t;i++)
        {
            scanf("%1d",&x);
            if(x==1)
            {
                if(vis[i])
                    n1--;
                else
                    n2++;
            }
        }
        cout<<max(n1,n2)<<endl;
    }
}

下面一种方法和第一种类似，不过用了两个数组来记录两次输入的状态，最后再循环一次仅记录相同下标不同数字的个数（分别记录0和1不同的个数）

#include <bits/stdc++.h>
using namespace std;
const int t = 1e5 + 10;
int b[t], c[t];
char a[t];
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    int x;
    while (n--)
    {
        memset(b, 0, sizeof(b));
        memset(c, 0, sizeof(c));
        cin >> x;
        for (int i = 0; i < x; i++)
        {
            cin >> a[i];
            if (a[i] == '1')
            {
                b[i]++;
            }
        }
        for (int i = 0; i < x; i++)
        {
            cin >> a[i];
            if (a[i] == '1')
            {
                c[i]++;
            }
        }
        int count1 = 0, count = 0, count2 = 0;
        for (int i = 0; i < x; i++)
        {
            if (b[i] != c[i])
            {
                if (b[i] == 1)
                {
                    count1++;
                }
                else
                {
                    count++;
                }
            }
        }
        int ret;
        if (count1 <= count)
        {
            cout << count << endl;
        }
        else
        {
            cout << count1 << endl;
        }
    }
    return 0;
}