#11：假日将尽的挣扎——6

cf818A

#include <bits/stdc++.h>
#define ll __int64
using namespace std;

ll n, k;

int main() {
    cin >> n >> k;
    ll a = n / 2 / (k+1);
    ll b = k * a;
    printf("%I64d %I64d %I64d
", a, b, n-a-b);
    return 0;
}

818B，没有指示的数字随便赋值即可。

#include <bits/stdc++.h>
#define ll __int64
using namespace std;

int n, m;
int l[105];
int a[105];
bool mark[105];
vector<int> v;

int main() {
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        cin >> l[i];
        int k = (l[i] - l[i-1] + n) % n;
        if (!k)    k = n;
        if (!a[l[i-1]]) {
            a[l[i-1]] = k;
        } else if (a[l[i-1]] != k) {
            puts("-1");
            return 0;
        }
    }

    for (int i = 1; i <= n; i++) {
        if (a[i]) {
            if (!mark[a[i]])
                mark[a[i]] = true;
            else {
                puts("-1");
                return 0;
            }
        } else {
            v.push_back(i);
        }
    }
    for (int i = 1; i <= n; i++) {
        if (!mark[i]) {
            a[v.back()] = i;
            v.pop_back();
        }
    }

    for_each(a+1, a+1+n, [](int x) {cout << x << " ";});
    return 0;
}

818C，sweep line，坐标点为下标差分计数。

#include <bits/stdc++.h>
using namespace std;

const int maxcor = 1e5 + 5;
int d, n, m;
int cntl, cntr, cntt, cntb;
struct Point {
    int x1, y1, x2, y2;
}sofa[maxcor];
int cnt_left[maxcor], cnt_right[maxcor], cnt_top[maxcor], cnt_bottom[maxcor];

inline bool l(int i) {
    int k = cnt_left[max(sofa[i].x1, sofa[i].x2) - 1];
    if (sofa[i].x1 != sofa[i].x2)    k--;
    return k == cntl;
}

inline bool r(int i) {
    int k = cnt_right[min(sofa[i].x1, sofa[i].x2) + 1];
    if (sofa[i].x1 != sofa[i].x2)    k--;
    return k == cntr;
}

inline bool t(int i) {
    int k = cnt_top[max(sofa[i].y1, sofa[i].y2) - 1];
    if (sofa[i].y1 != sofa[i].y2)    k--;
    return k == cntt;
}

inline bool b(int i) {
    int k = cnt_bottom[min(sofa[i].y1, sofa[i].y2) + 1];
    if (sofa[i].y1 != sofa[i].y2)    k--;
    return k == cntb;
}

int main() {
    cin >> d >> n >> m;
    for (int i = 1; i <= d; i++) {
        int x1, y1, x2, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        sofa[i] = {x1, y1, x2, y2};

        cnt_left[min(x1, x2)]++;
        cnt_right[max(x1, x2)]++;
        cnt_top[min(y1, y2)]++;
        cnt_bottom[max(y1, y2)]++;
    }
    cin >> cntl >> cntr >> cntt >> cntb;

    for (int i = 1; i <= 1e5; i++) {
        cnt_left[i] += cnt_left[i-1];
        cnt_top[i] += cnt_top[i-1];
    }
    for (int i = 1e5; i; i--) {
        cnt_right[i] += cnt_right[i+1];
        cnt_bottom[i] += cnt_bottom[i+1];
    }

    for (int i = 1; i <= d; i++) {
        if (l(i) && r(i) && t(i) && b(i)) {
            cout << i << endl;
            return 0;
        }
    }
    puts("-1");
    return 0;
}

818D，gameover之时将此数打标记，否则继续计数。

#include <bits/stdc++.h>
using namespace std;

const int maxcolor = 1e6 + 5;
bool mark[maxcolor];
int n, A, cntA;
int cnt[maxcolor];

int main() {
    cin >> n >> A;
    mark[A] = true;
    for (int i = 1; i <= n; i++) {
        int a;
        scanf("%d", &a);
        if (a == A)    cntA++;
        else {
            if (mark[a])
                continue;
            else if (cnt[a] < cntA)
                mark[a] = true;
            else    cnt[a]++;
        }
    }
    for (int i = 1; i <= 1e6; i++)
        if (!mark[i] && cnt[i] >= cntA) {
            cout << i << endl;
            return 0;
        }
    puts("-1");
    return 0;
}

BZOJ1011，鬼题勿做。思路根本就不是在求正解，以及迷之SPJ。

#include <cstdio>
#define db double

const int maxn = 1e5 + 5;
int n;
db A;
db m[maxn], sum[maxn];

int main() {
    scanf("%d%lf", &n, &A);
    for (int i = 1; i <= n; i++) {
        scanf("%lf", &m[i]);
        sum[i] = sum[i-1] + m[i];
    }
    for (int i = 1; i <= n; i++) {
        int k = int(A * db(i) + 1e-8);
        db ans = 0.0;
        if (k <= 100) {
            for (int j = 1; j <= k; j++)
                ans += m[i] * m[j] / (i-j);
        } else {
            ans = sum[k] * m[i] / (i-k/2);
        }
        printf("%.6lf
", ans);
    }
    return 0;
}

POJ3468，分块。

#include <cstdio>
#include <cmath>
#define ll long long
#define maxn 100005
#define rep(i, x, y) for (int i = x; i <= y; i++)

int n, m, t;
int L[maxn], R[maxn], pos[maxn];
ll a[maxn];
ll sum[maxn], add[maxn];

void Add(int l, int r, int k) {
    int p = pos[l], q = pos[r];
    if (p == q) {
        rep(i, l, r)    a[i] += k;
        sum[p] += k * (r - l + 1);
    } else {
        rep(i, p+1, q-1)    add[i] += k;
        rep(i, l, R[p])    a[i] += k;
        sum[p] += k * (R[p] - l + 1);
        rep(i, L[q], r)    a[i] += k;
        sum[q] += k * (r - L[q] + 1);
    }
}

ll query(int l, int r) {
    int p = pos[l], q = pos[r];
    ll ans = 0ll;
    if (p == q) {
        rep(i, l, r)    ans += a[i];
        ans += add[p] * (r - l + 1);
    } else {
        rep(i, p+1, q-1)    ans += sum[i] + add[i] * (R[i] - L[i] + 1);
        rep(i, l, R[p])    ans += a[i];
        ans += add[p] * (R[p] - l + 1);
        rep(i, L[q], r)    ans += a[i];
        ans += add[q] * (r - L[q] + 1);
    }
    return ans;
}

int main() {
    scanf("%d%d", &n, &m);
    rep(i, 1, n)    scanf("%lld", &a[i]);

    t = sqrt(n);
    rep(i, 1, t) {
        L[i] = (i-1) * t + 1;
        R[i] = i * t;
    }
    if (R[t] < n) {
        t++;
        L[t] = R[t-1] + 1;
        R[t] = n;
    }
    rep(i, 1, t) {
        rep(j, L[i], R[i]) {
            pos[j] = i;
            sum[i] += a[j];
        }
    }

    rep(i, 1, m) {
        char ch[3];
        int a, b;
        scanf("%s%d%d", ch, &a, &b);
        if (ch[0] == 'C') {
            int c;
            scanf("%d", &c);
            Add(a, b, c);
        } else {
            printf("%lld
", query(a, b));
        }
    }

    return 0;
}