2-1基础算法-枚举/模拟

1.枚举

[例1] 特别数的和

评测系统

#include 
using namespace std;
bool pa(int x) {
    while (x) {
        if (x % 10 == 2 || x % 10 == 1 || x % 10 == 0 || x % 10 == 9) {
            return true;
        }
        else {
            x = x / 10;
        }
    }
    return false;
}
int main()
{
    int sum=0;
    int m;
    cin >> m;
    for (int i = 1; i <= m; i++) {
        if (pa(i)) {
            sum += i;
        }
    }
    cout << sum;
    return 0;
}

[例2] 反序数

评测系统

#include 
using namespace std;
int main()
{
    int n;
    cin >> n;
    int a, b, c;
    cin >> a >> b >> c;
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        if (i % a != 0 && i % b != 0 && i % c != 0) {
            sum++;
        }
    }
    cout << sum;
    return 0;
}

[例3] 找到最多的数

评测系统

#include 
#include 
#include 
using namespace std;
int main()
{
    int n, m;
    cin >> n >> m;
    int num = n * m;
    map<int,int> mp;
    vector<int> a(num);//记录去重后的序列
    while (num--) {
        int x;
        cin >> x;
        if (!mp.count(x)) {
            a.push_back(x);
        }
        mp[x]++;
    }
    for (const auto& x:a) { //也可以不使用vector，每次从map中取出[x.y]
        if (mp[x] * 2 > n * m)
            cout << x;
    }
}

[例4]


评测系统

对于每种可能的颜色，我们尝试计算如果将走廊涂成这种颜色需要多少天。在遍历过程中要跳过已经是这种颜色的房子，每次移动k步。最后记录对于这种颜色所需的涂色天数，并与之前的最小天数进行比较，取较小值。
输入数据的最大值即为初始颜色种类数

#include 
#include 
using namespace std;
int main()
{
    int t;
    cin >> t;
    while (t--) {
        int n, k;
        cin >> n >> k;
        vector<int> a(n);
        int maxnum = 0;//记录初始颜色种类数
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
            maxnum=max(maxnum, a[i]);
        }
        int ans = n;
        for (int c = 1; c <= maxnum; ++c) {
            int cnt = 0;
            int idx = 0;
            while (idx < n) {
                if (a[idx] == c) {
                    idx++;
                    continue;
                }
                idx += k;
                cnt++;
            }
            ans = min(cnt, ans);
        }
        cout << ans << endl;
    }
}

2.模拟

[例1] 扫雷


评测系统

#include 
using namespace std;
const int N = 150;
int main()
{
    int n, m;
    cin >> n >> m;
    int a[N][N], ans[N][N];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cin >> a[i][j];
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (a[i][j] == 1) { 
                ans[i][j] = 9; 
                continue; 
            }
            int i2 = i, j2 = j;//遍历四周
            for (i2 = max(0, i - 1); i2 <= min(n - 1, i + 1); i2++) {
                for (j2 = max(0, j - 1); j2 <= min(m - 1, j + 1); j2++) {
                    if (a[i2][j2] == 1) ans[i][j]++;
                }
            }
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cout << ans[i][j]<<" ";
        }
        cout << endl;
    }
    return 0;
}

[例2] 灌溉


评测系统

使用类似于BFS的过程，用bool类型的irrigated数组记录当前位置是否已被灌溉，最终统计已被灌溉的数量。

#include 
#include 
using namespace std;
const int N = 200;
int main()
{
    int n, m, t;
    cin >> n >> m >> t;
    bool irrigated[N][N];
    queue<pair<int, int>> q;
    for (int i = 0; i < t; i++) {
        int r, c;
        cin >> r >> c;
        r--;
        c--;
        irrigated[r][c] = true;
        q.push({ r,c });
    }
    int k;
    cin >> k;
    for (int minute = 0; minute < k; ++minute) {
        int size = q.size();
        for (int i = 0; i < size; ++i) {
            pair<int,int > fr=q.front();//取到水管的位置坐标
            int x = fr.first;
            int y = fr.second;
            q.pop();
            for (int i2 = max(0,x - 1); i2 <= min(n,x + 1); i2++) { //四周灌溉
                for (int j2 = max(0,y - 1); j2 <= min(m,y + 1); j2++) {
                    if (!irrigated[i2][j2]) {
                        irrigated[i2][j2] = true;
                        q.push({i2, j2});
                    }
                }
            }
        }
    }
    int count = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (irrigated[i][j]) {
                count++;
            }
        }
    }
    cout << count;
    return 0;
}

也可使用map

#include 
#include 
using namespace std;
const int N = 200;
int main()
{
    int n, m, t;
    cin >> n >> m >> t;
    bool irrigated[N][N];
    map<int, int> q;
    for (int i = 0; i < t; i++) {
        int r, c;
        cin >> r >> c;
        r--;
        c--;
        irrigated[r][c] = true;
        q[r]=c;
    }
    int k;
    cin >> k;
    for (int minute = 0; minute < k; ++minute) {
        int size = q.size();
        for (const auto& i:q) {
            int x = i.first;
            int y = i.second;
            for (int i2 = max(0,x - 1); i2 <= min(n,x + 1); i2++) {
                for (int j2 = max(0,y - 1); j2 <= min(m,y + 1); j2++) {
                    if (!irrigated[i2][j2]) {
                        irrigated[i2][j2] = true;
                        q[i2]=j2;
                    }
                }
            }
        }
    }
    int count = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (irrigated[i][j]) {
                count++;
            }
        }
    }
    cout << count;
    return 0;
}

[例3] 回文日期

评测系统

#include 
#include 
using namespace std;
bool huiwen(string s) {
    if (s[0] == s[7] && s[1] == s[6] && s[2] == s[5] && s[3] == s[4]) {
        return true;
    }
    else
        return false;
}
bool er(string s) {
    if (s[0] == s[2] && s[2] == s[5] && s[5] == s[7] && s[1] == s[3] && s[3] == s[4] && s[4] == s[6]) {
        return true;
    }
    else
        return false;
}
int pdday(int year, int month) {
    if (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12)
        return 31;
    else if (month == 2) {
        if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0) {
            return 29;
        }
        else
            return 28;
    }
    else
        return 30;
}
string zifuchuan(int year, int month, int day) {
    string s=to_string(year);
    if (month / 10 == 0) {
        s += '0' + to_string(month);
    }
    else {
        s += to_string(month);
    }
    if (day / 10 == 0) {
        s += '0' + to_string(day);
    }
    else {
        s += to_string(day);
    }
    return s;
}
int main()
{
    int day, month, year;
    int chushi;
    int day1,day2;
    cin >> chushi;
    day1 = chushi % 10;
    chushi = chushi / 10;
    day2= chushi % 10;
    chushi = chushi / 10;
    day=day2 * 10 + day1;

    day1 = chushi % 10;
    chushi = chushi / 10;
    day2 = chushi % 10;
    chushi = chushi / 10;
    month = day2 * 10 + day1;

    year = chushi;
    bool flag = 0;
    bool pdhuiwen = 0;
    bool pder = 0;
    int i, j, k;
    for (i = year;; i++) {
        for (j = 1; j <= 12; j++) {
            if (flag == 0) {
                j = month;
            }
            for (k = 1; k <= pdday(year,month); k++) {
                if (flag == 0) {
                    k = day+1;
                    flag = 1;
                }
                string s=zifuchuan(i, j, k);
                if (pdhuiwen!=1&&huiwen(s)) {
                    cout << s << endl;
                    pdhuiwen = 1;
                }
                if (pder!=1&&er(s)) {
                    cout << s;
                    pder = 1;
                }
                if (pdhuiwen == 1 && pder == 1) {
                    return 0;
                }
            }
        }
    }
}

[例4] 小蓝和小桥的挑战

评测系统

#include 
#include 
#include 
using namespace std;
int main()
{
    int n;
    cin >> n;
    while (n--) {
        int m;
        cin >> m;
        vector<int> a(m);
        for(int i=0;i<m;i++){
            cin >> a[i];
        }
        int cnt = count(a.begin(), a.end(), 0); // #include 
        int sum = 0;
        for (const auto& x : a) {
            sum += x;
        }
        if (sum == 0) {
            cout << cnt + 1;
        }
        else
            cout << cnt<<endl;
    }
}

[例5] DNA序列修正

评测系统

#include 
using namespace std;
bool pipei(char a, char b) {
    return (a == 'A' && b == 'T') || (a == 'T' && b == 'A') || (a == 'C' && b == 'G') || (a == 'G' && b == 'C');
}
int main()
{
    int n;
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    int cnt = 0;//计数
    for (int i = 0; i < n; ++i) {
        if (!pipei(s1[i], s2[i])) {
            cnt++;
            for (int j = i + 1; j < n; j++) {
                if (!pipei(s1[j], s2[j])
                    && pipei(s1[i], s2[j])
                    && pipei(s1[j], s2[i])
                    ) {
                    swap(s2[i], s2[j]);
                    break;
                }
            }
        }
    }
    cout << cnt;
}

[例6] 无尽的石头


评测系统

#include 
using namespace std;
int ssum(int x) {
    int sum = 0;
    while (x) {
        sum += x % 10;
        x /= 10;
    }
    return sum;
}
int main()
{
    int t;
    cin >> t;
    while (t--) {
        int x = 0;
        cin >> x;
        int step = 0;
        bool end = 0;
        for (int i = 1; i <= x; i += ssum(i)) {
            if (i == x) {
                cout << step << endl;
                end = 1;
            }
            step++;
        }
        if (end == 0) {
            cout << -1 << endl;
        }
    }
}