HDU-4027 Can you answer these queries? (线段树 区间修改 区间查询)(The 36th ACM/ICPC Asia Regional Shanghai)
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Can you answer these queries?
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 20303 Accepted Submission(s): 4779
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.
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.
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
101 2 3 4 5 6 7 8 9 1050 1 101 1 101 1 50 5 81 4 8
Sample Output
Case #1:1976
Source
The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest
#include <stdio.h>#include <string.h>#include <algorithm>#include <vector>#include <math.h>using namespace std;#define maxn 100001long long c[maxn << 2];void build(int o, int l, int r, int id, long long x){if(l == r){c[o] = x; return;}int mid = l + r >> 1;if(id <= mid) build(o << 1, l, mid, id, x);if(id > mid) build(o << 1 | 1, mid + 1, r, id, x);c[o] = c[o << 1] + c[o << 1 | 1];}long long query(int o, int l, int r, int L, int R){if(l >= L && r <= R){return c[o];}long long ans = 0;int mid = l + r >> 1;if(mid >= L) ans += query(o << 1, l, mid, L, R);if(mid < R) ans += query(o << 1 | 1, mid + 1, r, L, R);return ans;}void add(int o, int l, int r, int L, int R){if(l == r){c[o] = sqrt(c[o]); return;}if(l >= L && r <= R && (c[o] == r - l + 1)) return;int mid = l + r >> 1;if(mid >= L) add(o << 1, l, mid, L, R);if(mid < R) add(o << 1 | 1, mid + 1, r, L, R);c[o] = c[o << 1] + c[o << 1 | 1];}int main(){int n, op, l, r, casenum = 1;long long x;while(scanf("%d", &n) != EOF){memset(c, 0, sizeof(c));for(int i = 1; i <= n; ++i){scanf("%lld", &x);build(1, 1, n, i, x);}int q;scanf("%d", &q);printf("Case #%d:\n", casenum++);while(q--){scanf("%d %d %d", &op, &l, &r);if(l > r) swap(l, r);if(op == 1) printf("%lld\n", query(1, 1, n, l, r));else add(1, 1, n, l, r);}puts("");}}/*题意:1e5的数,每次操作要么将区间内所有数开根号后取下整,要么查询区间和。思路:由于数字最大不超过1e19,最多开不超过10次根号,更新的时候可以暴力一点,查询区间和常规操作。注意给出的区间l,r可能l>r,出题人是真的贱。。。*/
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