題意  輸入n個數  然後有兩種操作   輸入0時將給定區間所有數都變為自己的開方   輸入1輸出給定區間所有數的和

雖然是區間更新  但每個點更新的不一樣  因此只能對單點進行更新  其實一個點最多被更新7次  2^64開平方7次後就變為1了  如果某個區間的數都變為了1  那麼對這個區間的開方就不用考慮了   另外要注意給你的區間可能是反的

#include <bits/stdc++.h>
#define lc p<<1,s,mid
#define rc p<<1|1,mid+1,e
#define mid ((s+e)>>1)
using namespace std;
typedef long long LL;
const int N = 100005;
LL sum[N * 4];

void pushup(int p)
{
    sum[p] = sum[p << 1] + sum[p << 1 | 1];
}

void build(int p, int s, int e)
{
    if(s == e)
    {
        scanf("%lld", &sum[p]);
        return;
    }
    build(lc);
    build(rc);
    pushup(p);
}

void update(int p, int s, int e, int l, int r)
{
    if(sum[p] == e - s + 1) return;  //[e,s]區間所有數都為1
    if(s == e)
    {
        sum[p] = sqrt(sum[p]);
        return;
    }
    if(l <= mid) update(lc, l, r);
    if(r > mid) update(rc, l, r);
    pushup(p);
}

LL query(int p, int s, int e, int l, int r)
{
    if(s == l && e == r) return sum[p];
    if(r <= mid) return query(lc, l, r);
    if(l > mid) return query(rc, l, r);
    return query(lc, l, mid) + query(rc, mid + 1, r);
}

int main()
{
    int n, m, a, b, c, k = 0;
    while(~scanf("%d", &n))
    {
        printf("Case #%d:\n", ++k);
        build(1, 1, n);
        scanf("%d", &m);
        while(m--)
        {
            scanf("%d%d%d", &c, &a, &b);
            if(b < a) swap(a, b);
            if(c) printf("%lld\n", query(1, 1, n, a, b));
            else update(1, 1, n, a, b);
        }
        puts("");
    }
    return 0;
}
 

Can you answer these queries?

Problem Description A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help.
You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line.

Notice that the square root operation should be rounded down to integer.
Input The input contains several test cases, terminated by EOF.
  For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000)
  The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 2⁶³.
  The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000)
  For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive.

Output For each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.
Sample Input 10 1 2 3 4 5 6 7 8 9 10 5 0 1 10 1 1 10 1 1 5 0 5 8 1 4 8
Sample Output Case #1: 19 7 6
Source