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  • Codeforces 671C Ultimate Weirdness of an Array 线段树 (看题解)

    Ultimate Weirdness of an Array

    写不出来, 日常好菜啊。。

    考虑枚举GCD, 算出一共有多少个对 f(l, r) <= GCD, 我们用fuc[ i ] 表示的是在 l = i 这个位置开始, 最小的合法的 R, 

    可以发现这个函数随 i 单调不下降, 枚举GCD 的时候, 找到GCD 的倍数的位置, 用线段树更新最大值。

    #include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 2e5 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 998244353;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}
    
    int n, a[N];
    int maxPos[N][2];
    int minPos[N][2];
    int cnt[N];
    LL H[N];
    int mx[2], mn[2], tot;
    
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
    struct segmentTree {
        LL sum[N << 2]; int lazy[N << 2], mn[N << 2], mx[N << 2];
        inline void pull(int rt) {
            sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
            mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
            mn[rt] = min(mn[rt << 1], mn[rt << 1 | 1]);
        }
        inline void push(int rt, int l, int r) {
            if(~lazy[rt]) {
                int mid = l + r >> 1;
                sum[rt << 1] = 1LL * (mid - l + 1) * lazy[rt];
                sum[rt << 1 | 1] = 1LL * (r - mid) * lazy[rt];
                lazy[rt << 1] = lazy[rt << 1 | 1] = lazy[rt];
                mx[rt << 1] = mn[rt << 1] = lazy[rt];
                mx[rt << 1 | 1] = mn[rt << 1 | 1] = lazy[rt];
                lazy[rt] = -1;
            }
        }
        void build(int l, int r, int rt) {
            lazy[rt] = -1;
            if(l == r) {
                sum[rt] = l;
                mx[rt] = l;
                mn[rt] = l;
                return;
            }
            int mid = l + r >> 1;
            build(lson); build(rson);
            pull(rt);
        }
        void update(int L, int R, int val, int l, int r, int rt) {
            if(R < l || r < L || R < L) return;
            if(mn[rt] >= val) return;
            if(L <= l && r <= R && mx[rt] <= val) {
                sum[rt] = 1LL * (r - l + 1) * val;
                mx[rt] = val;
                mn[rt] = val;
                lazy[rt] = val;
                return;
            }
            if(l == r) return;
            push(rt, l, r);
            int mid = l + r >> 1;
            update(L, R, val, lson);
            update(L, R, val, rson);
            pull(rt);
        }
    } Tree;
    
    int main() {
        memset(maxPos, 0xc0, sizeof(maxPos));
        memset(minPos, 0x3f, sizeof(minPos));
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            cnt[a[i]]++;
            if(i >= maxPos[a[i]][0]) {
                maxPos[a[i]][1] =  maxPos[a[i]][0];
                maxPos[a[i]][0] = i;
            } else if(i > maxPos[a[i]][1]) maxPos[a[i]][1] = i;
    
            if(i <= minPos[a[i]][0]) {
                minPos[a[i]][1] =  minPos[a[i]][0];
                minPos[a[i]][0] = i;
            } else if(i < minPos[a[i]][1]) minPos[a[i]][1] = i;
        }
        Tree.build(1, n, 1);
        for(int v = 200000; v >= 0; v--) {
            H[v] = 1LL * n * n - Tree.sum[1] + n;
            if(!v) break;
            mx[0] = mx[1] = -inf - 1;
            mn[0] = mn[1] = inf;
            tot = 0;
            for(int w = v; w <= 200000; w += v) {
                if(maxPos[w][0] == -inf - 1) continue;
    
                if(maxPos[w][0] >= mx[0]) mx[1] = mx[0], mx[0] = maxPos[w][0];
                else if(maxPos[w][0] > mx[1]) mx[1] = maxPos[w][0];
    
                if(maxPos[w][1] >= mx[0]) mx[1] = mx[0], mx[0] = maxPos[w][1];
                else if(maxPos[w][1] > mx[1]) mx[1] = maxPos[w][1];
    
    
                if(minPos[w][0] <= mn[0]) mn[1] = mn[0], mn[0] = minPos[w][0];
                else if(minPos[w][0] < mn[1]) mn[1] = minPos[w][0];
    
                if(minPos[w][1] <= mn[0]) mn[1] = mn[0], mn[0] = minPos[w][1];
                else if(minPos[w][1] < mn[1]) mn[1] = minPos[w][1];
    
                tot += cnt[w];
            }
            if(tot == 2) {
                int p1 = mn[0], p2 = mn[1];
                Tree.update(1, p1, p1, 1, n, 1);
                Tree.update(p1 + 1, p2, p2, 1, n, 1);
                Tree.update(p2 + 1, n, n + 1, 1, n, 1);
            } else if(tot == 3) {
                int p1 = mn[0], p2 = mn[1], p3 = mx[0];
                Tree.update(1, p1, p2, 1, n, 1);
                Tree.update(p1 + 1, p2, p3, 1, n, 1);
                Tree.update(p2 + 1, n, n + 1, 1, n, 1);
            } else if(tot > 3){
                int p1 = mn[0], p2 = mn[1], p3 = mx[1], p4 = mx[0];
                Tree.update(p2 + 1, n, n + 1, 1, n, 1);
                Tree.update(p1 + 1, p2, p4, 1, n, 1);
                Tree.update(1, p1, p3, 1, n, 1);
            }
        }
        LL ans = 0;
        for(int i = 1; i <= 200000; i++)
            ans += (H[i] - H[i - 1]) * i;
        printf("%lld
    ", ans);
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/10978952.html
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