传送门

L1-1 自动编程

#include<iostream>
#include<cstdio>

using namespace std;

int main(){
    int a;
    cin>>a;
    printf("print(%d)",a);
    return 0;
}

L1-2 太神奇了

#include<iostream>
#include<cstdio>

using namespace std;

int main(){
    int a,b;
    cin>>a>>b;
    cout<<(a+b)-1<<endl;
    return 0;
}

L1-3 洛希极限

#include<iostream>
#include<cstdio>

using namespace std;

int main(){
    double a, c;
    int b;
    cin>>a>>b>>c;
    if(b==0){
        double x=a*2.455;
        printf("%.2f ",x);
        if(x>=c)puts("T_T");
        else puts("^_^");
    }
    else {
        double x=a*1.26;
        printf("%.2f ",x);
        if(x<c)puts("^_^");
        else puts("T_T");
    }
    return 0;
}

L1-4 吃鱼还是吃肉

注意细节，包含所有情况

#include<iostream>
#include<cstdio>

using namespace std;

int main(){
    int T;
    cin>>T;
    while(T--){
        int a, b, c;
        cin>>a>>b>>c;
        if(a==1) {
            if (b == 130)printf("wan mei! ");
            else if (b < 130)printf("duo chi yu! ");
            else if (b > 130)printf("ni li hai! ");
            if (c == 27)printf("wan mei!\n");
            else if (c < 27)printf("duo chi rou!\n");
            else if (c > 27)printf("shao chi rou!\n");
        }
        else {
            if (b == 129)printf("wan mei! ");
            else if (b < 129)printf("duo chi yu! ");
            else if (b > 129)printf("ni li hai! ");
            if (c == 25)printf("wan mei!\n");
            else if (c < 25)printf("duo chi rou!\n");
            else if (c > 25)printf("shao chi rou!\n");
        }
    }
    return 0;
}

L1-5 不变初心数

我的做法是将数字转化为字符串，感觉比较好操作，数字转化为字符串to_string()，字符串转化为数字stoi()

#include<iostream>
#include<cstdio>

using namespace std;

int main() {
    int T;
    cin >> T;
    int ans, cnt;
    while (T--) {
        int n;
        cin >> n;
        string s = to_string(n);
        ans = 0;
        for (int i = 0; i < s.size(); i++) {
            ans += s[i] - '0';
        }
        bool flag = true;
        for (int i = 2; i <= 9; i++) {
            string s1 = to_string(n * i);
            //cout<<s1<<' ';
            cnt = 0;
            for (int j = 0; j < s1.size(); j++) {
                cnt += s1[j] - '0';
            }
//            cout<<cnt<<' ';cout<<endl;
            if (cnt != ans) {
                puts("NO");
                flag = false;
                break;
            }
        }
        if (flag)cout << ans << endl;
    }
    return 0;
}

L1-6 字母串

按照题目模拟即可

#include<iostream>
#include<cstdio>

using namespace std;

int main() {
    int T;
    cin >> T;
    int ans, cnt;
    while (T--) {
        string s;
        cin>>s;
        bool flag=true;
        for(int i=0;i<s.size();i++){
            if(i==s.size()-1)continue;
            if(s[i]>='A'&&s[i]<='Z'){
                if((s[i+1]+32)<'a'&&(s[i+1]+32)>'z'||(s[i+1]+1)<'A'&&(s[i+1]+1)>'Z')continue;
                else if(s[i+1]==(s[i]+32)||s[i+1]==(s[i]+1))continue;
                else flag=false;
                break;
            }
            else if(s[i]>='a'&&s[i]<='z'){
                if((s[i+1]-1)<'a'&&(s[i+1]-1)>'z'||(s[i+1]-32)<'A'&&(s[i+1]-32)>'Z')continue;
                else if(s[i+1]==(s[i]-32)||s[i+1]==(s[i]-1))continue;
                else flag=false;
                break;
            }
        }
        if(flag)puts("Y");
        else puts("N");
    }
    return 0;
}

L1-7 矩阵列平移

可以将偶数列单独提取出来，这样就变成了每次下一的位数就是列数%周期即可比较方便，先把奇数列的总和计算出来，最后再加上改变后的偶数列即可。

#include<bits/stdc++.h>

using namespace std;

const int N = 110;

int g[N][N];
int a[N], b[N];

int main() {
    int n, k, x;
    cin >> n >> k >> x;
    int t;
    for (int i = 1; i <= n; i++) {
        t = 0;
        for (int j = 1; j <= n; j++) {
            int m;
            cin >> m;
            if (j & 1)b[i] += m;
            else g[i][++t] = m;
        }
    }
    for (int i = 1; i <= t; i++) {
        int y = i % k;
        if (y == 0)y += k;
        for (int j = n - y; j >= 1; j--) {
            g[j + y][i] = g[j][i];
        }
        for (int j = 1; j <= y; j++)g[j][i] = x;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= t; j++) {
            b[i] += g[i][j];
        }
        if (i == 1)cout << b[i];
        else cout << ' ' << b[i];
    }
    return 0;
}

L1-8 均是素数

先把题目要求区间内的素数求出来，再写一个埃筛判断素数，最后用dfs来求出满足题目要求的素数组（因为数据范围较小），可以直接三层for循环暴力求；

#include<bits/stdc++.h>

using namespace std;

int ans;
int t=0;
int a[1010];
bool vis[1010];
int st[1000010];
int path[10];

bool check(int x) {
    st[0] = st[1] = 1;
    for (int i = 2; i * i <= x; i++) {
        if (!st[i]) {
            for (int j = i + i; j <= x; j += i) {
                st[j] = true;
            }
        }
    }
}

void dfs(int u) {
    if (u > 3) {
        int a1=path[1]*path[2]+path[3];
        int a2=path[2]*path[3]+path[1];
        int a3=path[3]*path[1]+path[2];
        if (!st[a1]&&!st[a2]&&!st[a3]&&path[1]<path[2]&&path[2]<path[3]&&path[1]<path[3]){
            //for(int i=1;i<=3;i++)cout<<path[i]<<' ';cout<<endl;
            ans++;
        }
            return;
    }
    for (int i = 1; i <= t; i++) {
        if (!vis[a[i]]) {
            vis[a[i]] = true;
            path[u] = a[i];
            dfs(u + 1);
            vis[a[i]] = false;
        }
    }
}

int main() {
    int l, r;
    cin >> l >> r;
    for (int i = l; i <= r; i++) {
        if (i == 1)continue;
        bool flag = true;
        for (int j = 2; j < i; j++) {
            if (i % j == 0) {
                flag = false;
                break;
            }
        }
        if (flag)a[++t] = i;
    }
    check(1010);
    dfs(1);
    cout << ans << endl;
    return 0;
}

L2-1 盲盒包装流水线

看懂题目，使用map里面涛哥vector即可

#include<bits/stdc++.h>

using namespace std;

const int N = 110;

vector<string> a;
vector<int> b;
int c[N];
int n, s;
map<string,vector<int>> pos;

int main() {
    cin >> n >> s;
    for (int i = 0; i < n; i++) {
        string s1;
        cin >> s1;
        a.push_back(s1);
        pos[s1].push_back(i);
    }
    for (int i = 0; i < n / s; i++) {
        for (int j = 0; j < s; j++)cin >> c[j];
        reverse(c, c + s);
        for (int j = 0; j < s; j++)b.push_back(c[j]);
        memset(c, 0, sizeof c);
    }
    //for (int i = 0; i < a.size(); i++)cout << a[i] << ' ' << b[i] << endl;
    int k;
    cin >> k;
    while (k--) {
        string x;
        cin >> x;
        if (pos.count(x)) {
            for (int i = 0; i < pos[x].size(); i++) {
                if (i == 0)cout << b[pos[x][i]];
                else cout << ' ' << b[pos[x][i]];
            }
            cout << endl;
        } else puts("Wrong Number");
    }
    return 0;
}

L2-2 点赞狂魔

好简单的模拟，简单结构体拍个序列即可

#include<bits/stdc++.h>

using namespace std;

struct S{
    string s;
    set<int> st;
    int cnt;
}c[110];

bool cmp(S a,S b){
    if(a.st.size()!=b.st.size())return a.st.size()>b.st.size();
    return a.cnt<b.cnt;
}

int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> c[i].s;
        int m;
        cin >> m;
        c[i].cnt = m;
        for (int j = 0; j < m; j++) {
            int x;
            cin >> x;
            c[i].st.insert(x);
        }
    }
    sort(c, c + n, cmp);
//    for (int i = 0; i < n; i++) {
//        cout << c[i].s << ' ' << c[i].st.size() << ' ' << c[i].cnt << endl;
//    }
    if (n == 0) {
        for (int i = 0; i < 3; i++) {
            if (i == 0)cout << '-';
            else cout << ' ' << '-';
        }
        cout << endl;
        return 0;
    }
    for (int i = 0; i < min(3,n); i++) {
        if (i == 0)cout << c[i].s;
        else cout << ' ' << c[i].s;
    }
    if (n < 3)
        for (int i = 0; i < 3 - n; i++) {
             cout << ' ' << '-';
        }
    cout << endl;
    return 0;
}

L2-3 浪漫侧影

先把数构建出来存到level数组里，一层一层的存，然后输出左右的视图，如果本层只有这一个元素，那么左右看到的都是它，如果有两个及其以上那么左边就是第一个，右边就是最后一个。

#include<bits/stdc++.h>

using namespace std;

const int N = 110;

int n;
int a[N], b[N], p[N];
vector<int> level[N];

void build(int al,int ar,int bl,int br,int d) {
    if (al > ar)return;
    int va = a[ar];
    int pos = p[va];
    level[d].push_back(va);
    build(al, al - bl + pos - 1, bl, pos - 1, d + 1);
    build(al - bl + pos, ar - 1, pos + 1, br - 1, d + 1);

}

int main() {
    cin >> n;
    for (int i = 0; i < n; i++)cin >> b[i], p[b[i]] = i;  //中序
    for (int i = 0; i < n; i++)cin >> a[i];  //后序
    build(0, n - 1, 0, n - 1, 0);
    cout << 'R' << ':' << ' ';
    for (int i = 0; level[i].size(); i++) {
        int k = level[i].size();
        if (i == 0) {
            if (level[i].size() == 1)cout << level[i][0];
            else cout << level[i][k - 1];
        } else {
            if (level[i].size() == 1)cout << ' ' << level[i][0];
            else cout << ' ' << level[i][k - 1];
        }
    }
    puts("");
    cout << 'L' << ':' << ' ';
    for (int i = 0; level[i].size(); i++) {
        if (i == 0)cout << level[i][0];
        else cout << ' ' << level[i][0];
    }
    return 0;
}

L2-4 哲哲打游戏

模拟模拟

#include<bits/stdc++.h>

using namespace std;

const int N = 100010;

vector<int> g[N];
int pos;
int n, m;
int file[110];

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        int k;
        cin >> k;
        while (k--) {
            int x;
            cin >> x;
            g[i].push_back(x);
        }
    }
    pos = 1;
//    for (int i = 1; i <= n; i++) {
//        for (auto j: g[i])
//            cout << j << ' ';
//        cout << endl;
//    }
    while (m--) {
        int a, b;
        cin >> a >> b;
        if (a == 0)pos = g[pos][b - 1];
        else if (a == 1) {
            cout << pos << endl;
            file[b] = pos;
        } else if (a == 2) {
            pos = file[b];
        }
    }
    cout << pos;
}

L3-2 拼题A打卡奖励

最经典的01背包的问题，但是这题为什么正常的做会超时，因为背包的维数太多了，所以超时，那么我们可以该改变一下思路，用价值作为背包内的状态，这样就不会超时了，那么我们每次转移就是求这个价值所需要的最小体积（min），最后枚举一下价值小于等于给定价值的就是对应的最小体积

#include<bits/stdc++.h>

using namespace std;

const int N = 1010, M = 30010;

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

inline int read()
{
    char c = getchar();int x = 0,s = 1;
    while(c < '0' || c > '9') {if(c == '-') s = -1;c = getchar();}//是符号
    while(c >= '0' && c <= '9') {x = x*10 + c -'0';c = getchar();}//是数字
    return x*s;
}


int main() {
    memset(f, 0x3f, sizeof f);
    n = read(), m = read();
    for (int i = 1; i <= n; i++)v[i] = read();
    for (int i = 1; i <= n; i++)w[i] = read();
    f[0] = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = M - 1; j >= w[i]; j--) {
            f[j] = min(f[j], f[j - w[i]] + v[i]);
        }
    }
    for (int i = M - 1; i >= 0; i--) {
        if (f[i] <= m) {
            cout << i << endl;
            return 0;
        }
    }
    return 0;
}