第十二届蓝桥杯省赛C++ B组题解

发表于 更新于 分类于 Valine：
本文字数： 488 阅读时长 ≈ 1 分钟

第十二届蓝桥杯省赛C++ B组题解

A.空间

思路

256MB=256 * 1024 * 1024 B

一个字节(Byte)等于8位(Bit)，所以一个32位二进制数占4B

代码

1
2
3
4
5
6
7
8
#include<bits/stdc++.h>
using namespace std;

int main()
{
    cout << 256 * 1024 * 1024 / 4 << endl;
    return 0;
}

答案

1
67108864

B.卡片

思路

用一个数组储存卡片个数，从1开始枚举，将每次枚举的数所需要的卡片分割出来，当储存卡片的数组小于零时，说明该数拼不出来，答案为上一个数。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
#include<bits/stdc++.h>

using namespace std;

int s[10];

bool check(int x)
{
    while (x)
    {
        int t = x % 10;
        x /= 10;
        if ( -- s[t] < 0) return false;
    }
    return true;
}

int main()
{
    for (int i = 0; i < 10; i ++ ) s[i] = 2021;

    for (int i = 1; ; i ++ )
        if (!check(i))
        {
            cout << i - 1 << endl;
            return 0;
        }
    return 0;
}

答案

1
3181

C.直线

思路

根据斜率判断不同的直线，当两条直线的k和b大于一个非常小的数(1e-8即$ 10^{-8} $)时，判断这两条线不是同一条直线。

k不存在的情况不枚举，最后加上二十条垂直x轴的线，x=0,1,2...19

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
#include<bits/stdc++.h>

using namespace std;

const int N = 200000;

int n;
struct Line
{
    double k, b;
    bool operator< (const Line& t) const
    {
        if (k != t.k) return k < t.k;
        return b < t.b;
    }
}l[N];

int main()
{
    for (int x1 = 0; x1 < 20; x1 ++ )
        for (int y1 = 0; y1 < 21; y1 ++ )
            for (int x2 = 0; x2 < 20; x2 ++ )
                for (int y2 = 0; y2 < 21; y2 ++ )
                    if (x1 != x2)
                    {
                        double k = (double)(y2 - y1) / (x2 - x1);
                        double b = y1 - k * x1;
                        l[n ++ ] = {k, b};
                    }

    sort(l, l + n);
    int res = 1;
    for (int i = 1; i < n; i ++ )
        if (fabs(l[i].k - l[i - 1].k) > 1e-8 || fabs(l[i].b - l[i - 1].b) > 1e-8)
            res ++ ;
    cout << res + 20 << endl;

    return 0;
}

答案

1
40257

D.货物摆放

思路

暴力枚举n的所有因数，三重循环枚举所有可能的情况，如果a*b*c==n，答案数++

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

int main()
{
    LL n;
    cin >> n;
    vector<LL> d;
    for (LL i = 1; i * i <= n; i ++ )
        if (n % i == 0)
        {
            d.push_back(i);
            if (n / i != i) d.push_back(n / i);
        }

    int res = 0;
    for (auto a: d)
        for (auto b: d)
            for (auto c: d)
                if (a * b * c == n)
                    res ++ ;
    cout << res << endl;

    return 0;
}

答案

1
2430

E.路径

思路

最短路径算法(spfa floyd dijkstra)

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
#include<bits/stdc++.h>

using namespace std;

const int N = 2200, M = N * 50;

int n;
int h[N], e[M], w[M], ne[M], idx;
int q[N], dist[N];
bool st[N];

int gcd(int a, int b)  // 欧几里得算法
{
    return b ? gcd(b, a % b) : a;
}

void add(int a, int b, int c)  // 添加一条边a->b，边权为c
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}

void spfa()  // 求1号点到n号点的最短路距离
{
    int hh = 0, tt = 0;
    memset(dist, 0x3f, sizeof dist);
    dist[1] = 0;
    q[tt ++ ] = 1;
    st[1] = true;

    while (hh != tt)
    {
        int t = q[hh ++ ];
        if (hh == N) hh = 0;
        st[t] = false;

        for (int i = h[t]; i != -1; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > dist[t] + w[i])
            {
                dist[j] = dist[t] + w[i];
                if (!st[j])     // 如果队列中已存在j，则不需要将j重复插入
                {
                    q[tt ++ ] = j;
                    if (tt == N) tt = 0;
                    st[j] = true;
                }
            }
        }
    }
}


int main()
{
    n = 2021;
    memset(h, -1, sizeof h);
    for (int i = 1; i <= n; i ++ )
        for (int j = max(1, i - 21); j <= min(n, i + 21); j ++ )
        {
            int d = gcd(i, j);
            add(i, j, i * j / d);
        }

    spfa();
    printf("%d\n", dist[n]);
    return 0;
}

答案

1
10266837

F.时间显示

思路

先将毫秒转换成秒，然后求小时(86400=60*60*24)、分钟(3600=60*60)、秒

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

int main()
{
    LL n;
    cin >> n;
    n /= 1000;
    n %= 86400;
    int h = n / 3600;
    n %= 3600;
    int m = n / 60;
    int s = n % 60;
    printf("%02d:%02d:%02d\n", h, m, s);
    return 0;
}

G.砝码称重

思路

dp
拓展P1441 砝码称重

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 110, M = 200010, B = M / 2;

int n, m;
int w[N];
bool f[N][M];

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ ) scanf("%d", &w[i]), m += w[i];

    f[0][B] = true;
    for (int i = 1; i <= n; i ++ )
        for (int j = -m; j <= m; j ++ )
        {
            f[i][j + B] = f[i - 1][j + B];
            if (j - w[i] >= -m) f[i][j + B] |= f[i - 1][j - w[i] + B];
            if (j + w[i] <= m) f[i][j + B] |= f[i - 1][j + w[i] + B];
        }

    int res = 0;
    for (int j = 1; j <= m; j ++ )
        if (f[n][j + B])
            res ++ ;
    printf("%d\n", res);
    return 0;
}

回溯法，超时

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
#include<bits/stdc++.h>

using namespace std;

int a[110];

int n;
set<int> st;

int back_trace(int size,int sum)
{
    //cout<<sum<<endl;
    if(sum!=0)
    st.insert(abs(sum));
    if(size==n) return 0;
    else
    {
        back_trace(size+1,sum+a[size]);
        back_trace(size+1,sum-a[size]);
        back_trace(size+1,sum);
    }
    return 0;
}
int main()
{
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    back_trace(0,0);
    cout<<st.size()<<endl;
    return 0;
}

H.杨辉三角形

思路

二分+找规律

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

int n;

LL C(int a, int b)
{
    LL res = 1;
    for (int i = a, j = 1; j <= b; i --, j ++ )
    {
        res = res * i / j;
        if (res > n) return res;
    }
    return res;
}

bool check(int k)
{
    LL l = k * 2, r = max((LL)n, l);
    while (l < r)
    {
        LL mid = l + r >> 1;
        if (C(mid, k) >= n) r = mid;
        else l = mid + 1;
    }
    if (C(r, k) != n) return false;

    cout << r * (r + 1) / 2 + k + 1 << endl;

    return true;
}

int main()
{
    cin >> n;
    for (int k = 16; ; k -- )
        if (check(k))
            break;
    return 0;
}

I.双向排序

思路

  1. 栈
  2. sort()暴力

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
#include<bits/stdc++.h>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 100010;

int n, m;
PII stk[N];
int ans[N];

int main()
{
    scanf("%d%d", &n, &m);
    int top = 0;
    while (m -- )
    {
        int p, q;
        scanf("%d%d", &p, &q);
        if (!p)
        {
            while (top && stk[top].x == 0) q = max(q, stk[top -- ].y);
            while (top >= 2 && stk[top - 1].y <= q) top -= 2;
            stk[ ++ top] = {0, q};
        }
        else if (top)
        {
            while (top && stk[top].x == 1) q = min(q, stk[top -- ].y);
            while (top >= 2 && stk[top - 1].y >= q) top -= 2;
            stk[ ++ top] = {1, q};
        }
    }
    int k = n, l = 1, r = n;
    for (int i = 1; i <= top; i ++ )
    {
        if (stk[i].x == 0)
            while (r > stk[i].y && l <= r) ans[r -- ] = k -- ;
        else
            while (l < stk[i].y && l <= r) ans[l ++ ] = k -- ;
        if (l > r) break;
    }
    if (top % 2)
        while (l <= r) ans[l ++ ] = k -- ;
    else
        while (l <= r) ans[r -- ] = k -- ;

    for (int i = 1; i <= n; i ++ )
        printf("%d ", ans[i]);
    return 0;
}

J.括号序列

思路

dp

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 5010, MOD = 1e9 + 7;

int n;
char str[N];
LL f[N][N];

LL work()
{
    memset(f, 0, sizeof f);
    f[0][0] = 1;
    for (int i = 1; i <= n; i ++ )
        if (str[i] == '(')
        {
            for (int j = 1; j <= n; j ++ )
                f[i][j] = f[i - 1][j - 1];
        }
        else
        {
            f[i][0] = (f[i - 1][0] + f[i - 1][1]) % MOD;
            for (int j = 1; j <= n; j ++ )
                f[i][j] = (f[i - 1][j + 1] + f[i][j - 1]) % MOD;
        }

    for (int i = 0; i <= n; i ++ )
        if (f[n][i])
            return f[n][i];
    return -1;
}

int main()
{
    scanf("%s", str + 1);
    n = strlen(str + 1);
    LL l = work();
    reverse(str + 1, str + n + 1);
    for (int i = 1; i <= n; i ++ )
        if (str[i] == '(') str[i] = ')';
        else str[i] = '(';
    LL r = work();
    printf("%lld\n", l * r % MOD);
    return 0;
}