Can you answer these queries?
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 18877 Accepted Submission(s): 4456
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.
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.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 +10;
long long n, m, p = 0;
struct xx{
long long l, r, v, f;
} T[N<<2];
void Pushup(int k){
T[k].v = T[k<<1].v + T[k<<1|1].v;
T[k].f = T[k<<1].f && T[k<<1|1].f;
}
void Build(long long l, long long r, long long k){
T[k].l = l, T[k].r = r, T[k].f = 0;
if(l == r){
scanf("%lld", &T[k].v);
return;
}
int mid = (l+r)>>1;
Build(l, mid, k<<1);
Build(mid+1, r, k<<1|1);
Pushup(k);
}
void Update(long long l, long long r, long long k){
if(T[k].l == T[k].r){
T[k].v = sqrt(T[k].v);
if(T[k].v == 1) T[k].f = 1;
return;
}
int mid = (T[k].l+T[k].r)>>1;
if(l <= mid && !T[k<<1].f) Update(l, r, k<<1);
if(r > mid && !T[k<<1|1].f) Update(l, r, k<<1|1);
Pushup(k);
}
long long ans;
void Query(long long l, long long r, long long k){
if(l <= T[k].l && r >= T[k].r){
ans += T[k].v;
return;
}
if(T[k].l == T[k].r) return;
int mid = (T[k].l+T[k].r)>>1;
if(l <= mid) Query(l, r, k<<1);
if(r > mid) Query(l, r, k<<1|1);
}
int main(){
while(scanf("%lld", &n) == 1){
Build(1, n, 1);
printf("Case #%lld:\n", ++p);
scanf("%lld", &m);
for(int i = 0; i < m; i++){
long long u, v, w;
scanf("%lld%lld%lld", &u, &v, &w);
if(w < v) swap(v, w);
if(!u){
Update(v, w, 1);
}
else{
ans = 0;
Query(v, w, 1);
printf("%lld\n", ans);
}
}
printf("\n");
}
}