HDU 1027 G - Can you answer these queries?

http://acm.hdu.edu.cn/showproblem.php?pid=4027

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 14057    Accepted Submission(s): 3264

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

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

 

虽然这个过了而且也写过题解。但是还是再写一次。

因为：

它输入的区间L和R大小可能不是固定的L < R，要手动判定一下。

那么我以后做题，也要这样。因为我遇过太多这些坑了。

无论数据有没有，我都判定一下

 

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#define lson L, mid, cur << 1
#define rson mid + 1, R, cur << 1 | 1
const int maxn = 100000 + 20;
LL sum[maxn << 2];
bool book[maxn << 2];
void pushUp(int cur) {
    sum[cur] = sum[cur << 1] + sum[cur << 1 | 1];
    book[cur] = book[cur << 1] && book[cur << 1 | 1];
}
void build(int L, int R, int cur) {
    book[cur] = 0;
    if (L == R) {
        cin >> sum[cur];
        return ;
    }
    int mid = (L + R) >> 1;
    build(lson);
    build(rson);
    pushUp(cur);
}
void upDate(int begin, int end, int L, int R, int cur) {
    if (book[cur]) return;
    if (L == R) {
        sum[cur] = sqrt(sum[cur]);
        if (sum[cur] == 1) book[cur] = 1;
        return;
    }
    int mid = (L + R) >> 1;
    if (begin <= mid) upDate(begin, end, lson);
    if (end > mid) upDate(begin, end, rson);
    pushUp(cur);
}
LL query(int begin, int end, int L, int R, int cur) {
    if (L >= begin && R <= end) {
        return sum[cur];
    }
    int mid = (L + R) >> 1;
    LL ans = 0;
    if (begin <= mid) ans += query(begin, end, lson);
    if (end > mid) ans += query(begin, end, rson);
    return ans;
}
int f;
int n;
void work() {
    printf("Case #%d:\n", ++f);
    build(1, n, 1);
    int q;
    cin >> q;
    for (int i = 1; i <= q; ++i) {
        int flag, L, R;
        scanf("%d%d%d", &flag, &L, &R);
        if (L > R) swap(L, R);
        if (flag == 0) upDate(L, R, 1, n, 1);
        else {
            cout << query(L, R, 1, n, 1) << endl;
        }
    }
}

int main() {
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    while (scanf("%d", &n) != EOF) {
        work();
        printf("\n");
    }
    return 0;
}