zoukankan      html  css  js  c++  java
  • Query on a string

    You have two strings SS and TT in all capitals.

    Now an efficient program is required to maintain a operation and support a query.

    The operation C~i~chC i ch with given integer ii and capital letter chch, changes the ii-th character of SSinto chch.

    The query Q~i~jQ i j asks the program to find out, in the substring of SS from the ii-th character to the jj-th one, the total number of TT appearing.

    Input Format

    The first line contains an integer TT, indicating that there are TT test cases.

    For each test case, the first line contains an integer N~(N leq 100000)N (N100000).

    The second line is the string S~(|S| leq 100000)S (S100000)and the third line is the string T~(|T| leq 10)T (T10).

    Each of the following NN lines provide a operation or a query as above descriptions.

    Output Format

    For each query, output an integer correponding to the answer.

    Output an empty line after each test case.

    样例输入

    1
    5
    AABBABA
    AA
    Q 1 3
    C 6 A
    Q 2 7
    C 2 B
    Q 1 5

    样例输出

    1
    2
    0
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<sstream>
    #include<algorithm>
    #include<queue>
    #include<deque>
    #include<iomanip>
    #include<vector>
    #include<cmath>
    #include<map>
    #include<stack>
    #include<set>
    #include<functional>
    #include<fstream>
    #include<memory>
    #include<list>
    #include<string>
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    
    const int MAXN = 100000 + 33;
    char s[MAXN];//字符串
    bool been[MAXN];//结尾记录是否匹配 匹配为1,不匹配为0
    char t[100];
    int T, n, a[MAXN];
    inline int lowbit(int x)
    {
        return x&(-x);
    }
    int getsum(int x)
    {
        if (x == 0) return 0;
        int sum = 0;
        while (x > 0)
        {
            sum += a[x];
            x -= x&(-x);
        }
        return sum;
    }
    void update(int x, int val)
    {
        while (x < MAXN)
        {
            a[x] += val;
            x += x&(-x);
        }
    }
    int main()
    {
        //ios::sync_with_stdio(0);
        scanf("%d", &T);
        int u, v;
        char op[2], to[2];
        while (T--)
        {
            memset(been, false, sizeof(been));
            memset(a, 0, sizeof(a));
            scanf("%d%s%s", &n, s + 1, t + 1);
            int l = strlen(t + 1), L = strlen(s + 1);
            for (int i = 1; i + l - 1 <= L; i++)
            {
                int cnt = 1;
                while (cnt <= l&&s[i + cnt - 1] == t[cnt])cnt++;
                if (cnt > l)
                {
                    update(i + l - 1, 1);
                    been[i + l - 1] = true;
                }
            }
            //for (int i = 1; i <= L; i++)
            //    cout << getsum(i) << endl;
            while (n--)
            {
                scanf("%s", op);
                if (op[0] == 'Q')
                {
                    scanf("%d%d", &u, &v);
                    printf("%d
    ", max(0, getsum(v) - getsum(u + l - 1- 1)));
                }
                else if(op[0] == 'C')
                {
                    scanf("%d%s", &u, to);
                    s[u] = to[0];
                    int len = max(1, u - l + 1);
                    for (int i = len; i <= u; i++)
                    {
                        if (i + l - 1> L) break;
                        int cnt = 1;
                        while (cnt <= l&&s[i + cnt - 1] == t[cnt])cnt++;
                        if (cnt > l)
                        {
                            if (been[i + l - 1]) continue;
                            been[i + l - 1] = 1;
                            update(i + l - 1, 1);
                        }
                        else if (been[i + l - 1])
                        {
                            been[i + l - 1] = 0;
                            update(i + l - 1, -1);
                        }
                    }
                }
            }
            cout << endl;
        }
    }
  • 相关阅读:
    django学习第85天Django的Ajax
    django学习第84天Django常用字段和参数
    django学习第83天Django聚合查询.分组查询.FQ查询
    django学习第82天Django多表查询
    django学习第81天Django模板层2(单表查询.模块的导入和继承.静态文件配置)
    django学习第80天Django模板层1
    django学习第79天Django视图层
    Linux 内核文档翻译
    Linux设备模型——设备驱动模型和sysfs文件系统解读
    内核空间内存申请函数kmalloc kzalloc vmalloc的区别
  • 原文地址:https://www.cnblogs.com/joeylee97/p/7515241.html
Copyright © 2011-2022 走看看