AtCoder Beginner Contest 087

目录

• A - Buying Sweets
• B - Coins
• C - Candies
• D - People on a Line

A - Buying Sweets

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int a, b, x;
int main() {
    cin >> x >> a >> b;
    x-=a;
    cout << x % b << endl;
    return 0;
}

B - Coins

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int a, b, c, x;
int main() {
    cin >> a >> b >> c >> x;
    int res = 0;
    for (int i = 0; i <= a; i++) {
        for (int j = 0; j <= b; j++) {
            for (int k = 0; k <= c; k++) {
                if (i * 500 + j * 100 + k * 50 == x) res++;
            }
        }
    }
    cout << res << endl;
    return 0;
}

C - Candies

给出一个2xN的矩阵，问从左上角走到右下角能获得的权值最大是多少

类似数塔模型的做法，推一下即可

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n, a[2][N], dp[2][N];
int main() {
    cin >> n;
    for (int i = 0; i < 2; i++) {
        for (int j = 0; j < n; j++) {
            cin >> a[i][j];
        }
    }
    for (int i = n - 1; i >= 0; i--) {
        dp[1][i] = dp[1][i + 1] + a[1][i];
        dp[0][i] = max(dp[1][i], dp[0][i + 1]) + a[0][i];
    }
    cout << dp[0][0] << endl;
    return 0;
}

D - People on a Line

给出n个人，他们都有一个权值，现在给出m个说法，每个说法代表第y个人比第x个人大d，问这m个说法是否能够自洽

带权并查集裸题

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n, m;
int f[N], d[N];
int findf(int x) {
    if (f[x] == x) return x;
    int tmp = findf(f[x]);
    d[x] += d[f[x]];
    return f[x] = tmp;
}
void Union(int x, int y, int w) {
    int fx = findf(x), fy = findf(y);
    if (fx != fy) {
        f[fy] = fx;
        d[fy] = w - d[y] + d[x];
    }
}

int main() {
    cin >> n>>m;
    for (int i = 0; i <= n; i++) f[i] = i;
    int flag = 0;
    for (int i = 0; i < m; i++) {
        int x, y, w;
        cin >> x >> y >> w;
        int fx = findf(x), fy = findf(y);
        if (fx == fy) {
            if (d[y] - d[x] != w) flag = 1;
        } else
            Union(x, y, w);
    }
    if (flag) cout << "No" << endl;
    else
        cout << "Yes" << endl;

    return 0;
}