codeforces round #433 div2

A:枚举一下就行了...居然wa了一发，题目一定要看清

#include<bits/stdc++.h>
using namespace std;
int n;
int main()
{
    cin >> n;
    int mid = n / 2 - ((n & 1) == 0);
    for(int i = n / 2; i; --i)
    {
        if(__gcd(i, n - i) == 1)
        {
            printf("%d %d
", i, n - i);
            return 0;
        }
    }
    return 0;
} 

B:这道题有点恶心，就是判断当前能不能每三个点有一个居民，判断一下就可以了

#include<bits/stdc++.h>
using namespace std;
int n, k;
int main()
{
    cin >> n >> k;
    if(k == 0)
    {
        puts("0 0");
        return 0;
    }
    if(n == k || k == 0) printf("0 ");
    else printf("1 ");
    int tmp1 = (n + 1) / 2, tmp2 = (n + 2) / 3, t = n % 3;
    if(k < tmp2) printf("%d
", k * 2);
    else printf("%d
", n - k);
    return 0;
} 

C:这种题只能贪心吧，似乎cf这种配对题都是贪心，开个set查询前继后继

#include<bits/stdc++.h>
using namespace std;
const int N = 300010;
struct data {
    long long c;
    int id;
    bool friend operator < (data a, data b)
    {
        return a.c > b.c;
    }
} a[N];
int n, k;
int answer[N];
set<long long> s;
int main()
{
    cin >> n >> k;
    for(int i = 1; i <= n; ++i) scanf("%lld", &a[i].c), a[i].id = i, s.insert(i + k);
    sort(a + 1, a + n + 1);
    long long ans = 0;
    for(int i = 1; i <= n; ++i)
    {
        long long t1 = *(s.lower_bound(a[i].id)), t2 = *(s.upper_bound(a[i].id)), tmp;
        if(abs(a[i].id - t1) < abs(a[i].id - t2)) tmp = t1;
        else tmp = t2;
        s.erase(tmp);
        ans += abs(a[i].id - tmp) * a[i].c;
        answer[a[i].id] = tmp;
    }
    cout << ans << endl;
    for(int i = 1; i <= n; ++i) printf("%d ", answer[i]);
    return 0;
} 

D:考试时想了会弃了...觉得大概是枚举一下...事实上枚举一下维护前缀后缀和就行了

#include<bits/stdc++.h>
using namespace std;
const int N = 1000010;
const long long inf = 1000000010; 
struct data {
    int pos, cost;
    data(int pos, int cost) : pos(pos), cost(cost) {}
};
vector<data> v1[N], v2[N];
int n, m, k, day, cnt;
int vis[N], can1[N], can2[N];
long long sum;
long long sum1[N], sum2[N], mn[N];
int main()
{
    cin >> n >> m >> k;
    for(int i = 1; i <= m; ++i)
    {
        int d, f, t, c;
        scanf("%d%d%d%d", &d, &f, &t, &c);
        day = max(day, d);
        if(t == 0) v1[d].push_back(data(f, c));
        else v2[d].push_back(data(t, c));
    }
    for(int i = 1; i <= n; ++i) mn[i] = inf, sum += inf;
    for(int i = 1; i <= day; ++i)
    {
        for(int j = 0; j < v1[i].size(); ++j) 
        {
            data tmp = v1[i][j];
            if(tmp.cost < mn[tmp.pos])
            {
                sum -= mn[tmp.pos];
                sum += tmp.cost;
                mn[tmp.pos] = tmp.cost;
            }
            if(vis[tmp.pos] == 0)
            {
                vis[tmp.pos] = 1;
                ++cnt;
            }
        }
        sum1[i] = sum;
        if(cnt == n) can1[i] = 1;
//        printf("cnt = %d sum1[%d] = %lld can1[%d] = %d
", cnt, i, sum1[i], i, can1[i]);
    }
    memset(vis, 0, sizeof(vis));
    sum = 0;
    cnt = 0;
    for(int i = 1; i <= n; ++i) mn[i] = inf, sum += inf; 
    for(int i = day; i ; --i)
    {
        for(int j = 0; j < v2[i].size(); ++j) 
        {
            data tmp = v2[i][j];
            if(tmp.cost < mn[tmp.pos])
            {
                sum -= mn[tmp.pos];
                sum += tmp.cost;                
                mn[tmp.pos] = tmp.cost;
            }
            if(vis[tmp.pos] == 0)
            {
                vis[tmp.pos] = 1;
                ++cnt;
            }    
        }
        sum2[i] = sum;
        if(cnt == n) can2[i] = 1;
//        printf("cnt = %d sum2[%d] = %lld can2[%d] = %d
", cnt, i, sum2[i], i, can2[i]);
    }
    long long ans = 1ll << 60;
//    puts("------------------------");
    for(int i = 1; i <= day - k + 1; ++i) 
    {
//        printf("can1[%d] = %d can2[%d] = %d
", i, can1[i], i + k + 1, can2[i + k + 1]);
        if(can1[i] && can2[i + k + 1]) ans = min(ans, sum1[i] + sum2[i + k + 1]);
    }    
    cout << ((ans == 1ll << 60) ? -1 : ans) << endl; 
    return 0;
}

E:考试时候石乐志觉得主席树爆空间...写了个莫队...然后错了...发现题目看错莫队做不了GG...其实很简单，静态查询区间点数,主席树维护一下，然后容斥做一下加减法就行了

#include<bits/stdc++.h>
using namespace std;
const int N = 200010;
int size[N * 30], lc[N * 30], rc[N * 30], root[N];
int n, q, cnt;
void update(int l, int r, int &x, int last, int pos)
{
    size[x = ++ cnt] = size[last] + 1;
    lc[x] = lc[last];
    rc[x] = rc[last];
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(pos <= mid) update(l, mid, lc[x], lc[last], pos);
    else update(mid + 1, r, rc[x], rc[last], pos);    
}
int query(int l, int r, int x, int y, int a, int b)
{
    if(l > b || r < a) return 0;
    if(l >= a && r <= b) return size[x] - size[y];
    int mid = (l + r) >> 1;
    return (query(l, mid, lc[x], lc[y], a, b) + query(mid + 1, r, rc[x], rc[y], a, b));
}
int main()
{
    cin >> n >> q;
    for(int i = 1; i <= n; ++i) 
    {
        int x;
        scanf("%d", &x); 
        update(1, n, root[i], root[i - 1], x);
    }
    while(q --)
    {
        int l, d, r, u;
        scanf("%d%d%d%d", &l, &d, &r, &u);
        long long ans = (long long)n * (long long)(n - 1) / 2ll, tmp;
        tmp = query(1, n, root[n], root[0], 1, d - 1);
        ans -= tmp * (tmp - 1) / 2ll;
        tmp = query(1, n, root[n], root[0], u + 1, n);
        ans -= tmp * (tmp - 1) / 2ll;
        tmp = query(1, n, root[l - 1], root[0], 1, n);
        ans -= tmp * (tmp - 1) / 2ll;
        tmp = query(1, n, root[n], root[r], 1, n);
        ans -= tmp * (tmp - 1) / 2ll;
        tmp = query(1, n, root[n], root[r], u + 1, n);
        ans += tmp * (tmp - 1) / 2ll;
        tmp = query(1, n, root[l - 1], root[0], u + 1, n);
        ans += tmp * (tmp - 1) / 2ll;
        tmp = query(1, n, root[n], root[r], 1, d - 1);
        ans += tmp * (tmp - 1) / 2ll;
        tmp = query(1, n, root[l - 1], root[0], 1, d - 1);
        ans += tmp * (tmp - 1) / 2ll;
        printf("%lld
", ans);    
    }
    return 0;
}