HDU 5443 The Water Problem (水题,暴力)
题意:给定 n 个数,然后有 q 个询问,问你每个区间的最大值。
析:数据很小,直接暴力即可,不会超时,也可以用RMQ算法。
代码如下:
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};
int n, m;
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
int main(){
int T;
cin >> T;
while(T--){
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
scanf("%d", &m);
while(m--){
int l, r;
scanf("%d %d", &l, &r);
int ans = -1;
for(int i = l; i <= r; ++i)
ans = max(ans, a[i]);
printf("%d\n", ans);
}
}
return 0;
}
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};
int n, m;
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int d[maxn][20]; void rmq_init(){
for(int j = 1; (1<<j) <= n; ++j)
for(int i = 1; i + (1<<j) - 1 <= n; ++i)
d[i][j] = max(d[i][j-1], d[i+(1<<(j-1))][j-1]);
} int rmq_query(int l, int r){
//int k = log(r-l+1)/log(2.0);
int k = 0;
while((1<<(k+1)) <= r-l+1) ++k;
return max(d[l][k], d[r-(1<<k)+1][k]);
} int main(){
int T; cin >> T;
while(T--){
scanf("%d", &n);
memset(d, 0, sizeof(d));
for(int i = 0; i < n; ++i)
scanf("%d", &d[i+1][0]);
rmq_init();
scanf("%d", &m);
while(m--){
int l, r;
scanf("%d %d", &l, &r);
printf("%d\n", rmq_query(l, r));
}
}
return 0;
}
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