AtCoder Beginner Contest 133

AtCoder Beginner Contest 133

https://atcoder.jp/contests/abc133

A - T or T

#include <bits/stdc++.h>

using namespace std;

int main (){
    int a, b, n;
    cin >> n >> a >> b;
    cout << min (a * n, b);
}#include <bits/stdc++.h>

using namespace std;

int main (){
    int a, b, n;
    cin >> n >> a >> b;
    cout << min (a * n, b);
}

B - Good Distance

#include <bits/stdc++.h>

using namespace std;
const int N = 15;
int a[N][N], n, m, ans;

int main (){
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> a[i][j];
        }
    }

    for (int i = 1; i < n; i++) {
        for (int ii = i + 1; ii <= n; ii++) {
            int sum = 0;
            for (int j = 1; j <= m; j++) {
                sum += (a[ii][j] - a[i][j]) * (a[ii][j] - a[i][j]);
            }
            if ((int)sqrt (sum) == sqrt(sum))   ans ++;
        }
    }
    cout << ans;
}

C - Remainder Minimization 2019

#include <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 15;
vector<int> v;
int ans = 2018 * 2018;

signed main (){
    int l, r;
    cin >> l >> r;
    for (int i = l; i < r; i++) {
        for (int j = i + 1; j <= r; j++) {
            ans = min (ans, (i * j) % 2019);
            if (!ans)   break;
        }
        if (!ans)   break;
    }
    cout << ans;
}

//找到离2019的倍数最近的几个

D - Rain Flows into Dams

#include <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 1e5 + 5;
int a[N], b[N], n;

signed main (){
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = n - 1; i >= 0; i -= 2)     b[0] +=a[i];
    for (int i = n - 2; i >= 0; i -= 2)     b[0] -=a[i];
    for (int i = 1; i < n; i++)     b[i] = 2 * a[i-1] - b[i-1];
    for (int i = 0; i < n; i++)     cout << b[i] << ' ';
}

//解方程

E - Virus Tree 2

#include <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 1e5 + 5, M = N * 2, mod = 1000000007;
int n, k, dep[N], cnt[N]; //深度, 深度为i的时候有多少个
int h[N], e[M], ne[M], idx;
int ans = 1, val[N];

void add (int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void dfs (int u, int fa) {
    dep[u] = dep[fa] + 1;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j == fa)    continue;
        dfs (j, u);
    }
}

void count (int u, int fa) {
    if (dep[u] == 2)    val[u] --;
    if (dep[u] >= 3)    val[u] -= 2;

    int flag = false, sum = 0;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j == fa)    continue;
        if (!flag) {
            flag = true;
            sum = val[j];
        }
        else {
            val[j] = sum - 1;
            sum = val[j];
        }
    }
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j == fa)    continue;
        count (j, u);
    }
}

signed main () {
    memset (h, -1, sizeof h);
    cin >> n >> k;
    for (int i = 1; i < n; i++) {
        int a, b;
        cin >> a >> b;
        add (a, b), add (b, a);
    }
    dfs (1, 0);
    for (int i = 1; i <= n; i++)    val[i] = k;
    count (1, 0);
    for (int i = 1; i <= n; i++) {
        ans = (ans * max (0ll, val[i])) % mod;
        if (!ans)   break;
    }
    
    cout << ans << endl;
}

//注意对0取max

F - Colorful Tree

树剖