第八届蓝桥杯省赛C++B组题解

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本文探讨了日期问题的排序与去重技巧，优化了购物清单计算方法，涉及等差素数列、承压计算、方格分割等算法实现。还介绍了如何使用编程处理日期、数位提取及最大公共子串问题。

总结

日期问题，注意判重和排序，搜索优化搜索方式，逆向思维。

购物单

方法1： 在excel中计算，复制粘贴后对数据进行以空格为标志分列，删除汉字，用公式计算。
方法2： 利用ctrl+F快捷键，将购物单复制到代码中，把**** 替换成+号，去掉汉字`折`，改为两位数 
         即9折改成90，在数字间加上 *,计算结果除以100即可，此方法不至于输入错误。

等差素数列

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 500010;

int primes[N], cnt;
bool st[N];

void get_primes(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] <= n / i; j ++ )
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    get_primes(N - 1);
    
    int k = 0;
    for (int i = 0; i < cnt; i ++ )
    {
        for (int j = 1; j <= 10000; j ++ )
        {
            for (k = 1; k < 10; k ++ )
            {
                if (st[primes[i] + k * j]) break;
            }
            if (k == 10)
            {
                cout << j << endl;
                return 0;
            }
        }
    }
    return 0;
}

承压计算

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 50;

int n = 30;
double g[N][N];
typedef long long LL;

int main()
{
    for (int i = 1; i < n; i ++ )
        for (int j = 1; j <= i; j ++ )
            cin >> g[i][j];
    
    for (int i = 1; i < n; i ++ )
        for (int j = 1; j <= i; j ++ )
        {
            g[i + 1][j] += g[i][j] / 2;
            g[i + 1][j + 1] += g[i][j] / 2;
        }
    
    double minn = 1e18, maxn = 0;
    for (int i = 1; i <= n; i ++ )
    {
        minn = min(minn, g[30][i]);
        maxn = max(maxn, g[30][i]);
    }
    
    printf("%.0lf", 2086458231 / minn * maxn);
    return 0;
}

方格分割

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 10;
bool st[N][N];
int ans = 0;

int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

void dfs(int x, int y)
{
    if (x == 0 || y == 0 || x == 6 || y == 6)
    {
        ans ++ ;
        return;
    }
    
    for (int i = 0; i < 4; i ++ )
    {
        int nx = x + dx[i], ny = y + dy[i];
        if (st[nx][ny]) continue;
        st[nx][ny] = true;
        st[6 - nx][6 - ny] = true;
        dfs(nx, ny);
        st[nx][ny] = false;
        st[6 - nx][6 - ny] = false;
    }
}

int main()
{
    st[3][3] = true;
    dfs(3,3);
    cout << ans / 4 << endl;
    return 0;
}

取数位

#include <stdio.h>
// 求x用10进制表示时的数位长度 
int len(int x){
    if(x<10) return 1;
    return len(x/10)+1;
}
    
// 取x的第k位数字
int f(int x, int k){
    if(len(x)-k==0) return x%10;
    return f(x/10,k);
}
    
int main()
{
    int x = 23574;
    printf("%d\n", f(x,3));
    printf("%d\n", f(893275,2));
}

最大公共子串

#include <stdio.h>
#include <string.h>

#define N 256
int f(const char* s1, const char* s2)
{
    int a[N][N];
    int len1 = strlen(s1);
    int len2 = strlen(s2);
    int i,j;
    
    memset(a,0,sizeof(int)*N*N);
    int max = 0;
    for(i=1; i<=len1; i++){
        for(j=1; j<=len2; j++){
            if(s1[i-1]==s2[j-1]) {
                a[i][j] = a[i - 1][j - 1] + 1;
                if(a[i][j] > max) max = a[i][j];
            }
        }
    }
    
    return max;
}

int main()
{
    printf("%d\n", f("abcdkkk", "baabcdadabc"));
    printf("%d\n", f("aaakkkabababa", "baabababcdadabc"));
    printf("%d\n", f("abccbaacbcca", "ccccbbbbbaaaa"));    
    printf("%d\n", f("abcd", "xyz"));
    printf("%d\n", f("ab", "ab"));
    return 0;
}

日期问题

这道题还需要排序、判重如果单独判断可能有点小麻烦，如果暴力遍历的就没有这些问题hh

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

int y, m, d;
int months[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int cnt;
struct Date
{
    int y, m, d;
    int date;
    
    bool operator< (const Date& t) const
    {
        if (y != t.y) return y < t.y;
        else if (m != t.m) return m < t.m;
        return d < t.d;
    }
}date[3];

bool is_yeap(int y)
{
    if (m != 2) return false;
    return y % 4 == 0 && y % 100 != 0 || y % 400 == 0;
}

bool check(int a, int b, int c)
{
    if (a < 60) y = a + 2000;
    else y = a + 1900;
    
    m = b, d = c;
    if (b < 1 || b > 12) return false;
    if (c < 1 || c > months[b] + is_yeap(y)) return false;
    return true;
}

int main()
{
    int a, b, c;
    scanf("%d/%d/%d", &a, &b, &c);
    
    
    if (check(a, b, c)) date[cnt ++ ] = {y, m, d, y*10000+m*100+d};
    if (check(c, a, b)) date[cnt ++ ] = {y, m, d, y*10000+m*100+d};
    if (check(c, b, a)) date[cnt ++ ] = {y, m, d, y*10000+m*100+d};

    sort(date, date + cnt);
    
    for (int i = 0; i < cnt; i ++ )
        if (i && date[i].date == date[i - 1].date) continue;
        else printf("%d-%02d-%02d\n", date[i].y, date[i].m, date[i].d);
    return 0;
}

包子凑数

先判断最大公约数如果>1无解，用DP存储不能凑到的数量，转化为完全背包问题

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 110, M = 10005;

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

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

int main()
{
    cin >> n;
    int d = 0;
    for (int i = 1; i <= n; i ++ ) 
    {
        cin >> w[i];
        d = gcd(d, w[i]);
    }
    
    if (d > 1) puts("INF");
    else
    {
        f[0] = true;
        for (int i = 1; i <= n; i ++ )
        {
            for (int j = w[i]; j <= M; j ++ )
                f[j] |= f[j - w[i]];
        }
        
        int res = 0;
        for (int i = 1; i < M; i ++ )
            res += f[i]==0;
        printf("%d\n", res);
    }
    
    return 0;
}

分巧克力

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 100010;
int n, k;
int h[N], w[N];

bool check(int mid)
{
    int cnt = 0;
    for (int i = 1; i <= n; i ++ )
    {
        cnt += (h[i] / mid) * (w[i] / mid);
        if (cnt >= k) return true;
    }
    return false;
}

int main()
{
    cin >> n >> k;
    for (int i = 1; i <= n; i ++ ) scanf("%d%d", &h[i], &w[i]);
    
    int l = 1, r = N;
    while (l < r)
    {
        int mid = l + r + 1>> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
    cout << r << endl;
    return 0;
}

k倍区间

枚举右端点，哈希记录左端点多少个右端点同%k

#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>

using namespace std;
typedef long long LL;

const int N = 100010;
int n, k;
LL s[N];
map<LL, int> mp;

int main()
{
    scanf("%d%d", &n, &k);
    
    mp[0] ++ ;
    LL res = 0;
    for (int i = 1; i <= n; i ++ )
    {
        scanf("%lld", &s[i]);
        s[i] = (s[i] + s[i - 1]) % k;
        
        res += mp[s[i]];
        mp[s[i]] ++ ;
    }
    printf("%lld\n", res);
    return 0;
}