Can you answer these queries?
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 19391 Accepted Submission(s): 4584
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.
/*
题意:
给你n个数,n<=100000,每个数不超过2^63
两种操作
0 x y 表示从x到y每个数都开方取下正
1 x y 表示从x到y每个数求和
题解:
2^63开方 6~7次变为1之后就不用操作了,也就是当sum==区间长度时候
这道题类似hdu5239平方后mod 9223372034707292160 即2 ^ 63 - 2 ^ 31
即任意数的平方MOD此数,重复操作至多29次就会进入一个不变的数。
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <cmath>
using namespace std;
typedef long long ll;
const int maxn = 101000*4;
const int INF = 0x3f3f3f3f;
struct node{
int l,r;
ll sum;
}tree[maxn];
void build(int node,int b,int e){
int mid = (b+e)/2;
tree[node].l = b;
tree[node].r = e;
if(b==e) {
scanf("%lld",&tree[node].sum);
return;
}
build(node*2,b,mid);
build(node*2+1,mid+1,e);
tree[node].sum = tree[node*2].sum + tree[node*2+1].sum;
}
void update(int node ,int ql,int qr){
if(tree[node].sum == (tree[node].r-tree[node].l+1)) return;
if(tree[node].l == tree[node].r){
tree[node].sum = (ll)sqrt(tree[node].sum);
return;
}
int mid = (tree[node].l + tree[node].r)/2;
if(ql<=mid) update(node*2,ql,qr);
if(qr>mid) update(node*2+1,ql,qr);
tree[node].sum = tree[node*2].sum + tree[node*2+1].sum;
}
ll query(int node,int ql,int qr){
ll ans1 = 0,ans2 = 0;
if(ql<=tree[node].l && qr>=tree[node].r)
return tree[node].sum;
int mid = (tree[node].l + tree[node].r)/2;
if(ql<=mid) ans1 = query(node*2,ql,qr);
if(qr>mid) ans2 = query(node*2+1,ql,qr);
return ans1+ans2;
}
int main(){
int n,m;
int icase = 1;
while(scanf("%d",&n)!=EOF){
printf("Case #%d:\n",icase++);
build(1,1,n);
scanf("%d",&m);
while(m--){
int t,x,y;
scanf("%d%d%d",&t,&x,&y);
if(x>y) swap(x,y);///这里有坑
if(t==0) update(1,x,y);
if(t==1) printf("%lld\n",query(1,x,y));
}
puts("");
}
return 0;
}