cf618 训练cde

面完试第一场vp cf，感觉题目挺规矩的，可能是有些历史了

c

#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
int sum[N];
int main() {
    int n;
    scanf("%d", &n);
    vector<int> a(n);
    int ma = 0;
    for (auto& i : a) {
        scanf("%d", &i);
        int cnt = 0, t = i;
        while (t) {
            if (t & 1) {
                ++sum[cnt];
            }
            t >>= 1;
            cnt++;
        }
        ma = max(ma, cnt);
    }
    int ans = inf;
    for (int i = ma - 1; i >= 0; i--) {
        if (sum[i] == 1) {
            ans = i;
            break;
        }
    }
    if (ans != inf) {
        int tp = 0;
        for (int i = 0; i < n; i++) {
            if (a[i] & (1 << ans)) {
                tp = i;
                break;
            }
        }
        printf("%d", a[tp]);
        for (int i = 0; i < n; i++) {
            if (i != tp) {
                printf(" %d", a[i]);
            }
        }
        printf("
");
    }
    else {
        for (int i = 0; i < n; i++) {
            printf("%d", a[i]);
            if (i != n - 1) printf(" ");
            else printf("
");
        }
    }
    return 0;
}

d,有点几何的感觉

#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
const double  eps = 1e-9;
struct point {
    int x, y;
    point operator - (const point& b) {
        return { x - b.x, y - b.y };
    }
    bool operator == (const point& b) {
        return x == b.x && y == b.y;
    }
};
bool ki(point a, point b, point c, point d) {
    return (b - a) == (c - d);
}
int main() {
    int n;
    scanf("%d", &n);
    vector<point> q(n);
    for (auto& v : q) {
        scanf("%d%d", &v.x, &v.y);
    }
    if (n & 1) {
        printf("NO
");
    }
    else {
        bool can = true;
        for (int i = 0; can && i < n / 2; i++) {
            if (!ki(q[i], q[i + 1], q[i + n / 2], q[(i + n / 2 + 1) % n])) {
                can = false;
            }
        }
        if (can)printf("YES
");
        else printf("NO
");
    }
    return 0;
}

e,也是几何。。

#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
const double  eps = 1e-9;
struct point {
    int x, y;
    point operator - (const point& b) {
        return { x - b.x, y - b.y };
    }
    bool operator == (const point& b) {
        return x == b.x && y == b.y;
    }
};
bool ki(point a, point b, point c, point d) {
    return (b - a) == (c - d);
}
int main() {
    int n;
    scanf("%d", &n);
    vector<point> q(n);
    for (auto& v : q) {
        scanf("%d%d", &v.x, &v.y);
    }
    if (n & 1) {
        printf("NO
");
    }
    else {
        bool can = true;
        for (int i = 0; can && i < n / 2; i++) {
            if (!ki(q[i], q[i + 1], q[i + n / 2], q[(i + n / 2 + 1) % n])) {
                can = false;
            }
        }
        if (can)printf("YES
");
        else printf("NO
");
    }
    return 0;
}