省赛组队赛暨省赛积分赛1

省赛组队赛暨省赛积分赛1

来写总结。自闭了我就是垃圾，越训越烂www
https://vjudge.net/contest/553737#overview
先把补了的题放上，剩下的明天看看能不能写。

D. Dasha and Very Difficult Problem

贪心构造，也是我唯一过的一题。

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

using namespace std;
typedef pair<int, int> pii;
const int N = 1e5 + 5;
int n, L, R;
int a[N], l[N], r[N], ans[N];
pii c[N];

void solve () {
    scanf ("%d%d%d", &n, &L, &R);
    for (int i = 1; i <= n; i++) {
        scanf ("%d", &a[i]);
        l[i] = L - a[i], r[i] = R - a[i];
    }
    for (int i = 1; i <= n; i++)    scanf ("%d", &c[i].first), c[i].second = i;
    //c取最小的且比上一个大即可
    sort (c + 1, c + n + 1);
    int lst = 0;
    for (int i = 1; i <= n; i++) {
        int id = c[i].second, b = max (L, a[id] + lst + 1);

        ans[id] = b;
        lst = b - a[id];
    }

    int minn = ans[1], maxn = ans[1];
    for (int i = 1; i <= n; i++)    minn = min (minn, ans[i]), maxn = max (maxn, ans[i]);
    if (maxn - minn > R - L) {
        cout << -1;
        return ;
    }
    int dx = minn - L;
    for (int i = 1; i <= n; i++)    cout << ans[i] - dx << ' ';
}

int main () {
    solve ();
}

C. Chessboard

这个真的好可惜，当时急了，只想着最暴力的做法，明明那个n为奇数都已经标粗了还是没细想！！！还以为是状压之类的赛后补提发现只有两种最终局面，所以直接暴力check即可。。。真的心态出大问题。。。服了我自己

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

using namespace std;
const int N = 105, M = 205;
int a[5][N][N], n;
ll ans;
int seq[] = {0, 1, 2, 3};

void solve () {
    //check两种局面
    do {
        //0开头(0,3是0开头偶0; 1,2是1开头偶1)
        //1开头(0,3是1开头偶1; 1,2是0开头偶0)
        ll cnt1 = 0, cnt2 = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                int k = (i + j) & 1;
                if (k == 0) { //偶
                    if (a[seq[0]][i][j] != 0)    cnt1 ++;
                    else    cnt2 ++;
                    if (a[seq[1]][i][j] != 1)    cnt1 ++;
                    else    cnt2 ++;
                    if (a[seq[2]][i][j] != 1)    cnt1 ++;
                    else    cnt2 ++;
                    if (a[seq[3]][i][j] != 0)    cnt1 ++;
                    else    cnt2 ++;
                }
                else {
                    if (a[seq[0]][i][j] != 1)    cnt1 ++;
                    else    cnt2 ++;
                    if (a[seq[1]][i][j] != 0)    cnt1 ++;
                    else    cnt2 ++;
                    if (a[seq[2]][i][j] != 0)    cnt1 ++;
                    else    cnt2 ++;
                    if (a[seq[3]][i][j] != 1)    cnt1 ++;
                    else    cnt2 ++;
                }
            }
        }
        ans = min (ans, min (cnt1, cnt2));
    } while (next_permutation (seq, seq + 4));
}

int main () {
    cin >> n;
    ans = 4 * n * n + 1;
    for (int k = 0; k < 4; k++) {
        for (int i = 1; i <= n; i++) {
            string s;
            cin >> s;
            for (int j = 1; j <= n; j++) {
                a[k][i][j] = s[j-1] - '0';
            }
        }
        //check内部有无冲突
    }
    solve ();
    cout << ans << endl;
}

//其实终局是固定的，只有两种可能(奇数决定了0/1开头之后的局面是固定的)，只需枚举判断记录最小值

J. The Parade

贪心，但是没贪对。
我原有的错误思路：先把整段的放了，然后剩余的和上一段拼凑。
实际上真正的贪心，本段先和上一段剩下来的拼凑，余数再继承。
这里面要细想一下逻辑，最好画个图辅助理解。。
也是最后把自己绕进去了还想了个dp啥的。。。

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

using namespace std;
const int N = 30005;
ll n, k, a[N];

bool check (ll x) { //关键在于怎么放
    //同一行中所有士兵的身高差不能超过 1
    ll cnt = 0, res = 0; //看行数是否>=k, res为上行的遗留
    for (int i = 1; i <= n; i++) { //先放小的，然后调整空间更多一些?
        if (a[i] + res < x) {
            res = a[i] % x;
        }
        else {
            cnt += (a[i] + res) / x;
            res = (a[i] + res) % x;
        } 
    }
    return cnt >= k;
}

void solve () {
    scanf ("%lld%lld", &n, &k);
    ll l = 0, r = 0, sum = 0;
    for (int i = 1; i <= n; i++)    scanf ("%lld", &a[i]), sum += a[i];
    //二分该单最大数量
    r = sum / k;
    while (l < r) {
        ll mid = l + r + 1 >> 1;
        if (check (mid))    l = mid;
        else    r = mid - 1;
    }
    printf ("%lld\n", l * k);
}

int main () {
    int t;
    scanf ("%d", &t);
    while (t --)    solve ();
}

特别注意答案是有可能为0的！！！！