BZOJ 3211: 花神游历各国

Description

Input

Output

每次x=1时,每行一个整数,表示这次旅行的开心度

Sample Input

4

1 100 5 5

5

1 1 2

2 1 2

1 1 2

2 2 3

1 1 4

Sample Output

101

11

11

HINT

对于100%的数据, n ≤ 100000,m≤200000 ,data[i]非负且小于10^9

Solution

双倍经验题!同BZOJ3038。传送门->

因为10^9顶多开5~6次方就成了1了(当然这里的等于是向下取整的)因此对于修改操作,如果某一段不是1或不是0,就暴力修改,如果是1/0就不管他。修改完之后update一下就好了。

Code

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#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;
const int INF = 1009999999;
int n;
LL sum[500005], A[500005];
bool all[500005];
void pu(int now) {
    sum[now] = sum[now << 1] + sum[now << 1 | 1];
    all[now] = all[now << 1] && all[now << 1 | 1];
}
void build(int now, int l, int r) {
    if (l == r) {
        sum[now] = A[l];
        if (sum[now] == 1 || sum[now] == 0)
            all[now] = true;
        return;
    }
    int mid = (l + r) >> 1;
    build(now << 1, l, mid);
    build(now << 1 | 1, mid + 1, r);
    pu(now);
}
LL query(int now, int l, int r, int ll, int rr) {
    if (ll <= l && r <= rr) {
        return sum[now];
    }
    int mid = (l + r) >> 1;
    LL ret = 0;
    if (ll <= mid) ret += query(now << 1, l, mid, ll, rr);
    if (rr > mid) ret += query(now << 1 | 1, mid + 1, r, ll, rr);
    return ret;
}
void change(int now, int l, int r, int ll, int rr) {
    if (all[now]) {
        return;
    }
    if (l == r) {
        sum[now] = sqrt(sum[now]);
        if (sum[now] == 1 || sum[now] == 0)
            all[now] = true;
        return;
    }
    int mid = (l + r) >> 1;
    if (ll <= mid) change(now << 1, l, mid, ll, rr);
    if (rr > mid) change(now << 1 | 1, mid + 1, r, ll, rr);
    pu(now);
}
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
        scanf("%lld", A+i);
    build(1, 1, n);
    int m, x, y, z;
    scanf("%d", &m);
    while (m--) {
        scanf("%d%d%d", &x, &y, &z);
        if (y > z) swap(y, z);
        if (x==1) {
            printf("%lld\n", query(1, 1, n, y, z));
        } else {
            change(1, 1, n, y, z);
        }
    }
    return 0;
}