【BZOJ3211】花神游历各国(线段树)

Description


Solution

线段树维护,由于有效开根最多 log 次,直接暴力开根即可。


Code

/**************************************
 * Au: Hany01
 * Prob: [BZOJ3211] 花神游历各国
 * Date: Jul 31st, 2018
 * Email: [email protected]
**************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef vector<int> VI;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = j; i < i##_end_; ++ i)
#define For(i, j ,k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define SZ(a) ((int)(a.size()))
#define ALL(a) a.begin(), a.end()
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define x first
#define y second
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define y1 wozenmezhemecaia 
#ifdef hany01
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...)
#endif

template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

template<typename T> inline T read() {
    register char c_; register T _, __;
    for (_ = 0, __ = 1, c_ = getchar(); !isdigit(c_); c_ = getchar()) if (c_ == '-')  __ = -1;
    for ( ; isdigit(c_); c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 1e5 + 5;

LL sm[maxn << 2], mx[maxn << 2];

#define lc (t << 1)
#define rc (lc | 1)
#define mid ((l + r) >> 1)

inline void maintain(int t) { sm[t] = sm[lc] + sm[rc], mx[t] = max(mx[lc], mx[rc]); }

void build(int t, int l, int r) {
    if (l == r) sm[t] = mx[t] = read<LL>();
    else build(lc, l, mid), build(rc, mid + 1, r), maintain(t);
}

inline void update(int t, int l, int r, int x, int y) {
    if (l == r) { sm[t] = mx[t] = sqrt(mx[t]); return; }
    if (mx[t] <= 1) return;
    if (x <= mid) update(lc, l, mid, x, y);
    if (y >  mid) update(rc, mid + 1, r, x, y);
    maintain(t);
}

inline LL query(int t, int l, int r, int x, int y) {
    if (x <= l && r <= y) return sm[t];
    if (y <= mid) return query(lc, l, mid, x, y);
    if (x >  mid) return query(rc, mid + 1, r, x, y);
    return query(lc, l, mid, x, y) + query(rc, mid + 1, r, x, y);
}

int main()
{
#ifdef hany01
    File("bzoj3211");
#endif

    static int n = read<int>();
    build(1, 1, n);
    for (static int q = read<int>(), x, l, r; q --; ) {
        x = read<int>(), l = read<int>(), r = read<int>();
        if (l > r) swap(l, r);
        if (x & 1) printf("%lld\n", query(1, 1, n, l, r));
        else update(1, 1, n, l, r);
    }

    return 0;
}

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转载自blog.csdn.net/hhaannyyii/article/details/81321240