2017 雷火研发笔试3.20

总共4个编程题，分别是10，20,30,40分。

水题AC代码：

#include "iostream"
#include "string.h"
#include "vector"

using namespace std;

int n, m;

int v[10005];

void turn(int l, int r)
{
    for (int i = l; i <= r; i++)
    {
        v[i] = !v[i];
    }
}

int main()
{
    while (cin >> n >> m)
    {
        memset(v, 0, 10005);
        int a, b;
        for (int i = 0; i < m; i++)
        {
            cin >> a >> b;
            turn(a, b);
        }
        
        for (int i = 1; i <= n; i++)
            cout << v[i];

        cout << endl;
    }
}

//5 2
//1 2
//2 4

第二题：

第三题：

并查集，秒A

#include "iostream"
#include "string.h"
using namespace std;

int  pre[105];
bool t[105];               //t 用于标记独立块的根结点

int Find(int x)
{
    int r = x;
    while (r != pre[r])
        r = pre[r];

    int i = x, j;
    while (pre[i] != r)
    {
        j = pre[i];
        pre[i] = r;
        i = j;
    }
    return r;
}

void mix(int x, int y)
{
    int fx = Find(x), fy = Find(y);
    if (fx != fy)
    {
        pre[fy] = fx;
    }
}

int main()
{
    int N, M, a, b, i, j, ans;
    while (cin>>N>>M)
    {
        for (i = 1; i <= N; i++)          //初始化 
            pre[i] = i;

        for (i = 1; i <= M; i++)          //吸收并整理数据 
        {
            cin >> a >> b;
            mix(a, b);
        }

        memset(t, 0, sizeof(t));
        for (i = 1; i <= N; i++)          //标记根结点
        {
            t[Find(i)] = 1;
        }
        for (ans = 0, i = 1; i <= N; i++)
            if (t[i])
                ans++;

        cout<<ans - 1<<endl;

    }
    return 0;
}

第四题：

规约成构造回文串的题，A了81%

#include "iostream"
#include "string"
#include "vector"
#include "algorithm"
#include "set"
using namespace std;

int tt = 0;
int helper(string s)
{
    string rev = s;
    reverse(rev.begin(), rev.end());

    if (rev == s)    return 0;

    int size = s.size();
    vector<vector<int>> c(size + 1, vector<int>(size + 1));
    for (int i = 1; i <= size; ++i)
        for (int j = 1; j <= size; ++j)
        {
            if (s[i - 1] == rev[j - 1])
            {
                c[i][j] = c[i - 1][j - 1] + 1;
            }
            else if (c[i - 1][j] >= c[i][j - 1])
            {
                c[i][j] = c[i - 1][j];
            }
            else if (c[i - 1][j] < c[i][j - 1])
            {
                c[i][j] = c[i][j - 1];
            }
        }
    return size - c[size][size];
}
int main()
{
    string s;
    while (cin >> s)
    {
        set<char> se;
        for (int i = 0; i < s.length(); i++)
        {
            se.insert(s[i]);
        }
        tt = se.size();
        if (helper(s) == 0)
            cout << tt << endl;
        else
            cout << helper(s) +tt -1 << endl;
    }
        
}

//0.81