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bzoj3211 花神游历各国

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=3211

【题解】

区间开根号,由于每个数被开根号不会很多次就变成1,每次我们暴力开根下去,同时记录s[x]表示x这个区间内是不是全是1,如果是就不用开下去了

这样保证了每个数最多被开不多次。

技术分享
# include <math.h>
# include <stdio.h>
# include <string.h>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n;
int seq[M];

ll w[M]; 
bool s[M]; 
inline void build(int x, int l, int r) {
    s[x] = 0; 
    if(l == r) {
        w[x] = seq[l];
        return ;
    }
    int mid = l+r>>1;
    build(x<<1, l, mid);
    build(x<<1|1, mid+1, r);
    w[x] = w[x<<1] + w[x<<1|1];
}

inline void change(int x, int l, int r, int L, int R) {
    if(s[x]) return; 
    if(l == r) {
        w[x] = sqrt(w[x]);
        if(w[x] <= 1) s[x] = 1; 
        return ;
    }
    int mid = l+r>>1;
    if(R <= mid) change(x<<1, l, mid, L, R); 
    else if(L > mid) change(x<<1|1, mid+1, r, L, R);
    else change(x<<1, l, mid, L, mid), change(x<<1|1, mid+1, r, mid+1, R);
    w[x] = w[x<<1] + w[x<<1|1];
    s[x] = s[x<<1] & s[x<<1|1]; 
}

inline ll query(int x, int l, int r, int L, int R) {
    if(L <= l && r <= R) return w[x];
    int mid = l+r>>1;
    if(R <= mid) return query(x<<1, l, mid, L, R);
    else if(L > mid) return query(x<<1|1, mid+1, r, L, R);
    else return query(x<<1, l, mid, L, mid) + query(x<<1|1, mid+1, r, mid+1, R); 
}

int main() {
    
    scanf("%d", &n); 
    for (int i=1; i<=n; ++i) scanf("%d", seq+i); 
    build(1, 1, n);
    int q; scanf("%d", &q);
    while(q--) {
        int opt, l, r;
        scanf("%d%d%d", &opt, &l, &r); if(l>r) swap(l, r); 
        if(opt == 1) {
            printf("%lld\n", query(1, 1, n, l, r));
        } else change(1, 1, n, l, r);
    }

    return 0;
}
View Code

 

bzoj3211 花神游历各国