HDU 5443 The Water Problem 解题报告（如题）

最新推荐文章于 2020-08-23 13:49:15 发布

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最新推荐文章于 2020-08-23 13:49:15 发布

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解决水源查询问题，通过给定一系列水源大小，在多个查询中快速找出指定范围内最大水源的大小。

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The Water Problem

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 222 Accepted Submission(s): 181

Problem Description

In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with

a1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing

2 integers

l and

r , please find out the biggest water source between

al and

ar .

Input

First you are given an integer

T(T≤10) indicating the number of test cases. For each test case, there is a number

n(0≤n≤1000) on a line representing the number of water sources.

n integers follow, respectively

a1,a2,a3,...,an , and each integer is in

{1,...,106} . On the next line, there is a number

q(0≤q≤1000) representing the number of queries. After that, there will be

q lines with two integers

l and

r(1≤l≤r≤n) indicating the range of which you should find out the biggest water source.

Output

For each query, output an integer representing the size of the biggest water source.

Sample Input

Sample Output

Source

2015 ACM/ICPC Asia Regional Changchun Online

解题报告：其实一开始让我写这题的解题报告我是拒绝的。

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <functional>
#include <cassert>
#include <bitset>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define ff(i, n) for(int i=0,END=(n);i<END;i++)
#define fff(i, n, m) for(int i=(n),END=(m);i<=END;i++)
#define dff(i, n, m) for(int i=(n),END=(m);i>=END;i--)
#define travel(e, u) for(int e=first[u], v=vv[first[u]]; ~e; e=nxt[e])
#define mid ((l+r)/2)
#define bit(n) (1ll<<(n))
#define clr(a, b) memset(a, b, sizeof(a))
#define debug(x) cout << #x << " = " << x << endl;

#define ls (rt << 1)
#define rs (ls | 1)
#define lson l, m, ls
#define rson m + 1, r, rs

void work();

int main() {
    work();
    return 0;
}

/**************************Beautiful GEGE**********************************/

int a[1111];

void work() {
    int T; scanf("%d", &T);
    fff(cas, 1, T) {
        int n; scanf("%d", &n);
        ff(i, n) scanf("%d", a + i);

        int q; scanf("%d", &q);
        ff(i, q) {
            int l, r; scanf("%d%d", &l, &r);
            printf("%d\n", *max_element(a + l - 1, a + r));
        }
    }
}