codeforces Round 314 div.2

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这次比赛是用小号打的，于是现在小号比大号rating高。。。QwQ

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A题和B题很简单不多说。

C题：统计子序列中长度为3，公比为 k 的等比序列的个数。

从后往前处理，分别统计满足条件长度为1,2,3的序列的个数，
f(i,j)=∑f(i−1,t)|t=j∗k，实现时离散化一下就好~

然而，当时只判断了 a[i]*k<=a[max] ，而没有判断 a[i]*k>=a[min] (k可能为负数)， 所以FST了。。。

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const int maxn = 2e5 + 5, size = maxn*3;int n, k, a[maxn];long long ans;int cnt[maxn], p[maxn], pl;long long f[maxn], g[maxn];int main(){#ifndef ONLINE_JUDGE    freopen("C.in","r",stdin);    freopen("C.out","w",stdout);#endif    read(n), read(k);    for(int i = 1; i <= n; i++)        read(a[i]), p[i] = a[i];    std::sort(p + 1, p + n + 1);    pl = std::unique(p + 1, p + n + 1) - (p + 1);    for(int i = n; i > 0; i--)    {        int t = std::lower_bound(p + 1, p + pl + 1, a[i]) - p;        if((long long)a[i]*k <= p[pl] && (long long)a[i]*k >= p[1])        {            int v = std::lower_bound(p + 1, p + pl + 1, a[i]*k) - p;            if(a[i]*k == p[v]) g[t] += f[v], f[t] += cnt[v];        }        cnt[t]++;    }    for(int i = 1; i <= pl; i++) ans += g[i];    write(ans);#ifndef ONLINE_JUDGE    fclose(stdin);    fclose(stdout);#endif    return 0;}

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D题：从结束状态逆向处理，用并查集维护最长区间，直到满足可放船数>=实际船数才得到答案。

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const int maxn = 2e5 + 5, maxk = maxn;int n, k, a, m;int x[maxk], cnt;bool e[maxn];int fa[maxn], size[maxn];int find(int x){    return (x == fa[x])?x:(fa[x] = find(fa[x]));}void gather(int x,int y){    if(x > y) std::swap(x, y);    x = find(x), y = find(y), fa[y] = x;    if(x != y) size[x] += size[y], size[y] = 0;}void init(){    read(n), read(k), read(a), read(m);    for(int i = 1; i <= m; i++)        read(x[i]), e[x[i]] = true;}void prework(){    for(int i = 1; i <= n; i++) fa[i] = i, size[i] = 1;    for(int i = 1; i <= n; i++)        if(i > 1 && !e[i] && !e[i-1])            gather(i, i - 1);    for(int i = 1; i <= n; i++)        if(fa[i] == i && !e[i]) cnt += (size[i]+1)/(a+1);       }int solve(){    if(cnt >= k) return -1;    for(int i = m; i >= 1; i--)    {        if(x[i] > 1 && !e[x[i]-1])        {            cnt -= (size[find(x[i]-1)]+1)/(a+1);            gather(x[i] - 1, x[i]);        }        if(x[i] < n && !e[x[i]+1])        {            cnt -= (size[find(x[i]+1)]+1)/(a+1);            gather(x[i], x[i] + 1);        }        cnt += (size[find(x[i])]+1)/(a+1);        e[x[i]] = false;        if(cnt >= k) return i;    }    return 0;}int main(){#ifndef ONLINE_JUDGE    freopen("D.in","r",stdin);    freopen("D.out","w",stdout);#endif    init(), prework();    write(solve());#ifndef ONLINE_JUDGE    fclose(stdin);    fclose(stdout);#endif    return 0;}