南外集训 Day 8

The eighth day of intensive training at Nanjing Foreign Language School.
第一题，模拟

#include <bits/stdc++.h>
using namespace std;
int a[17], ans = 4, cnt, tot;
int main() {
    for (int i = 1; i <= 16; i++) {
        cin >> a[i];
    }
    for (int i = 1; i <= 4; i++) {
        cnt = 0;
        tot = 0;
        for (int j = 1; j <= 16; j++) {
            if (a[j] > cnt) {
                cnt = a[j];
                tot = j;
            }
        }
        a[tot] = -1;
    }
    for (int i = 1; i <= 16; i++) {
        if (abs(cnt - a[i]) < 300) {
            ans++;
        }
    }
    cout << ans;
    return 0;
}

第二题，数学题

#include <bits/stdc++.h>
using namespace std;
int a[17], ans = 4, cnt, tot;
int main() {
    for (int i = 1; i <= 16; i++) {
        cin >> a[i];
    }
    for (int i = 1; i <= 4; i++) {
        cnt = 0;
        tot = 0;
        for (int j = 1; j <= 16; j++) {
            if (a[j] > cnt) {
                cnt = a[j];
                tot = j;
            }
        }
        a[tot] = -1;
    }
    for (int i = 1; i <= 16; i++) {
        if (abs(cnt - a[i]) < 300) {
            ans++;
        }
    }
    cout << ans;
    return 0;
}

第三题，循环枚举寻找最大公因数

#include <bits/stdc++.h>
using namespace std;
long long n, x, p[2020020], c[2020020], tot = 2147483647;
bool b = false;
int main() {
    cin >> n >> x;
    for (int i = 1; i <= n; i++) {
        cin >> p[i];
        c[i] = abs(x - p[i]);
        tot = min(c[i], tot);
    }
    while (tot > 1) {
        b = true;
        for (int i = 1; i <= n; i++) {
            if (c[i] % tot != 0) {
                b = false;
                break;
            }
        }
        if (b == true) {
            break;
        } else {
            tot--;
        }
    }
    cout << tot;
    return 0;
}

第四题，递归合并

#include <bits/stdc++.h>
using namespace std;
int n, A, B, C, l[999], ans = 0x3f3f3f3f;
void dfs(int q, int x, int y, int z, int p) {
    if (x && y && z) {
        ans = min(ans, abs(x - A) + abs(y - B) + abs(z - C) + p - 30);
    }
    if (q > n) {
        return;
    }
    dfs(q + 1, x + l[q], y, z, p + 10);
    dfs(q + 1, x, y + l[q], z, p + 10);
    dfs(q + 1, x, y, z + l[q], p + 10);
    dfs(q + 1, x, y, z, p);
}
int main() {
    cin >> n >> A >> B >> C;
    for (int i = 1; i <= n; i++) {
        cin >> l[i];
    }
    dfs(0, 0, 0, 0, 0);
    cout << ans;
    return 0;
}