• 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
    ", 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
    ", rmq_query(l, r));
            }
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/5722120.html
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