The third "intelligence Cup" National College Students' IT skills competition (solution to group B of the preliminary competition)

Game connection

Course registration

Ideas

Sign in problem ：
It can be simulated directly .

AC Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
void solve() {
    
    int n, v, m, a;
    int cnt  = 0, ans = 0;
    scanf("%d%d%d%d", &n, &v, &m, &a);
    for (int i = 1; i <= n; ++i) {
    
        cnt++;
        ans += v;
        if (cnt == m) {
    
            v += a;
            cnt = 0;
        }
    }
    printf("%d\n", ans);
}
int main () {
    
    solve();
    return 0;
}

Final exam results

Ideas

Or simulation

AC Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
void solve() {
    
    double x;
    scanf("%lf", &x);
    double ans = 0;
    if (x >= 90) {
    
        ans = 4.0;
    } else if (x >= 60 && x < 90) {
    
        ans = 4.0 - (90 - x) * 0.1;
    } else {
    
        int y = sqrt(x) * 10;
        x = y;
        if (x >= 60 && x < 90) {
    
            ans = 4.0 - (90 - x) * 0.1;
        } else {
    
            ans = 0.0;
        }
    }

    printf("%0.1f\n", ans);
}
int main () {
    
    solve();
    return 0;
}

volunteer

Ideas

Simple order , Custom collation .

AC Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 5e5 + 5;
struct Node {
    
   int t, k, ind;
}node[MAXN];
bool cmp(Node a, Node b) {
    
    if (a.k * a.t == b.k * b.t) {
    
        if (a.k == b.k) {
    
            return a.ind < b.ind;
        }
        else {
    
             return a.t > b.t;
        }

    }
    return a.k * a.t > b.k * b.t;
}
void solve() {
    
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
    
        scanf("%d%d", &node[i].t, &node[i].k);
        node[i].ind = i;
    }
    sort(node + 1, node + n + 1, cmp);
    for (int i = 1; i <= n; ++i) {
    
        if (i == n) {
    
            printf("%d\n", node[i].ind);
        } else {
    
            printf("%d ", node[i].ind);
        }

    }
}
int main () {
    
    solve();
    return 0;
}

terminal

Ideas

Use two map Make mapping relationship , Take time to map file names , File name demapping time , One-to-one correspondence . Then carry out four operations .

AC Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
map<string, int> gt;
map<int, string> gf;
void solve() {
    
    int n, cnt = 0;
    cin >> n;
    string f1, s1;
    while(n--) {
    
        cin >> f1;
        if (f1[0] != 'l') {
    
            cin >> s1;
            if (f1 == "touch") {
    
                if (!gt.count(s1)) {
    
                    gt[s1] = ++cnt;
                    gf[cnt] = s1;
                }
            } else if (f1 == "rename") {
    
                string yyy;
                cin >> yyy;
                if (gt.count(s1) && gt.count(yyy) == 0) {
    
                    int t = gt[s1];
                    gf[t] = yyy;
                    gt.erase(s1);
                    gt[yyy] = t;
                }
            } else if (f1 == "rm"){
    
                int t = gt[s1];
                gt.erase(s1);
                gf.erase(t);
            }
        } else {
    
            for (map<int,string>::iterator it = gf.begin(); it!=gf.end(); ++it) {
    
                cout << (it->second) << "\n";
            }
        }
    }
}
int main () {
    
    solve();
    return 0;
}

luck

Ideas

Simple search , enumeration n The situation of each bit , The maximum number of recursion is 6¹⁰ ≈ 6e⁷, Recursion will not explode , And there is no need to take a mold , The biggest case is 6¹⁰.

AC Code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
ll n, k, sum = 0;
void dfs(int cnt, ll num) {
    
    if (cnt == n) {
    
        if (num % k == 0) {
    
            sum++;
        }
        return;
    }
    for (int i = 1; i <= 6; ++i) {
    
        dfs(cnt + 1, num * 10 + i);
    }
}
void solve() {
    
    scanf("%lld%lld", &n, &k);
    dfs(0 , 0);
    printf("%lld\n", sum);
}
int main () {
    
    solve();
    return 0;
}