智算之道2020第一场初赛题解报告

T1

#include 
using namespace std;

const int N = 100005;
bool a[N];

int main()
{
    int n, m;
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
    {
        a[i] = 1;
    }

    int t;
    for(int i = 1; i <= m; i++)
    {
        cin >> t;
        for(int i = 1; i <= n; i++)
        {
            if(i % t == 0)
            {
                a[i] = 0;
            }
        }
    }

    int cnt = 0;
    for(int i = 1; i <= n; i++)
    {
        cnt += a[i];
    }

    cout << cnt;

    return 0;
}

T2

解法一：枚举，60分

#include 
using namespace std;

string change(string s)
{
    int len = s.size();
    int pos;
    for(int i = len - 1; i >= 0; i--)
    {
        if(s[i] == '1')
        {
            pos = i;
            break;
        }
    }

    for(int i = 0; i <= pos; i++)
    {
        if('1' == s[i])
        {
            s[i] = '0';
        }
        else
        {
            s[i] = '1';
        }
    }

    return s;
}

int main()
{
    //freopen("T2.in", "r", stdin);

    int n;
    string s;
    cin >> n >> s;
    string target;
    for(int i = 0; i < n; i++)
    {
        target.push_back('0');
    }

    int cnt = 0;
    while(s != target)
    {
        s = change(s);
        cnt++;
    }

    cout << cnt;
    return 0;
}

解法二

#include
#include
#include
using namespace std;

typedef long long LL;
const int maxn = 2e5+10;
int a[maxn];

int main()
{
    int n;
	string s;
	cin >> n >> s;

	int cnt = 0;
	char lastC = s[0];
	for(int i = 1; i < s.length(); i++)
	{
		if(lastC != s[i])
		{
			lastC = s[i];
			cnt++;
		}
	}

	if(lastC == '1')
    {
        cnt++;
    }

	cout << cnt;

	return 0;
}

T3

解法一：枚举，30分

#include 

using namespace std;

int main()
{
    //freopen("T3.in", "r", stdin);

    string s, t;
    cin >> s >> t;
    sort(s.begin(), s.end());

    int cnt = 0;
    do
    {
        if(string::npos != t.find(s))
        {
            cnt++;
        }
    }while(next_permutation(s.begin(), s.end()));

    cout << cnt;

    return 0;
}

解法二：尺取法+map去重

#include
using namespace std;

int main()
{
    //freopen("T3.in", "r", stdin);

    int cnt = 0;
    string s, t, subT;
    cin >> s >> t;

    map<string, bool> vis; //判断字符串是否被遍历过
    int cnt1[123]={0}, cnt2[123]={0};//a~z的ASCII码是97~122

    int sLen = s.length(), tLen = t.length();
    for(int i=0; i<sLen; i++)
    {
        cnt1[s[i]]++;
    }

    for(int startPos=0; startPos <= tLen - sLen; startPos++)
    {
        if(startPos==0)
        {
            for(int j=0; j < sLen; j++)
            {
                cnt2[t[j]]++;
            }
        }
        else
        {
            //整体往右挪动一个位置，长度仍为sLen
            cnt2[t[startPos - 1]]--;
            cnt2[t[startPos + sLen - 1]]++;
        }

        bool same = true;//s各字符数量是否与t的子串各字符数量一样
        for(int j='a'; j<='z'; j++)
        {
            if(cnt1[j] != cnt2[j])
            {
                same=false;
            }
        }

        //subT是t中长度为sLen的子串
        subT = t.substr(startPos, sLen);
        if(same && !vis[subT])
        {
            vis[subT]=true;
            cnt++;
        }
    }

    cout << cnt;

    return 0;
}

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