• hdu 5057 Argestes and Sequence


    Argestes and Sequence

    Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 511    Accepted Submission(s): 127


    Problem Description
    Argestes has a lot of hobbies and likes solving query problems especially. One day Argestes came up with such a problem. You are given a sequence a consisting of N nonnegative integers, a[1],a[2],...,a[n].Then there are M operation on the sequence.An operation can be one of the following:
    S X Y: you should set the value of a[x] to y(in other words perform an assignment a[x]=y).
    Q L R D P: among [L, R], L and R are the index of the sequence, how many numbers that the Dth digit of the numbers is P.
    Note: The 1st digit of a number is the least significant digit.
     

    Input
    In the first line there is an integer T , indicates the number of test cases.
    For each case, the first line contains two numbers N and M.The second line contains N integers, separated by space: a[1],a[2],...,a[n]—initial value of array elements.
    Each of the next M lines begins with a character type.
    If type==S,there will be two integers more in the line: X,Y.
    If type==Q,there will be four integers more in the line: L R D P.

    [Technical Specification]
    1<=T<= 50
    1<=N, M<=100000
    0<=a[i]<=$2^{31}$ - 1
    1<=X<=N
    0<=Y<=$2^{31}$ - 1
    1<=L<=R<=N
    1<=D<=10
    0<=P<=9
     

    Output
    For each operation Q, output a line contains the answer.
     

    Sample Input
    1 5 7 10 11 12 13 14 Q 1 5 2 1 Q 1 5 1 0 Q 1 5 1 1 Q 1 5 3 0 Q 1 5 3 1 S 1 100 Q 1 5 3 1
     

    Sample Output
    5 1 1 5 0 1
     

    Source
     

    题解:

           这道题有三种版本号的 题解,本来题目不难,就是限制空间:1.分块算法解决,2.离线树状数组,3.卡空间的树状数组

    这里先介绍第一种算法:

           学习了一下分块算法,事实上还蛮简单的,就是将n组元素分成m组,每组合并成一块,查询时,仅仅要看元素在那几块,相加即可了。

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    using namespace std;
    
    struct Block
    {
        int nt[10][10];
    }block[400];
    int num[100010];
    
    int cal(int d)
    {
        int ans=1;
        for(int i=1;i<=d;i++)
        {
            ans*=10;
        }
        return ans;
    }
    
    int init(int n)
    {
        int s=(int)sqrt((double)n),t=0;
        int m=n/s+1;
    
        memset(block,0,sizeof(block));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
            s=i/m;t=num[i];
            for(int j=0;j<=9;j++)
            {
                block[s].nt[j][t%10]++;
                t/=10;
            }
        }
        return m;
    }
    
    void work(int k,int n,int m)
    {
        char s[2];
        int l,r,d,p,tl,tr,td,tp,ans=0;
        while(m--)
        {
            scanf("%s",s);
            if(s[0]=='S')
            {
                scanf("%d%d",&d,&p);
                td=d;td/=k;
                for(int j=0;j<=9;j++)
                {
                    block[td].nt[j][num[d]%10]--;
                    num[d]/=10;
                }
                num[d]=p;tp=p;
                for(int j=0;j<=9;j++)
                {
                    block[td].nt[j][tp%10]++;
                    tp/=10;
                }
            }
            else
            {
               ans=0;
               scanf("%d%d%d%d",&l,&r,&d,&p);
               tl=l;tl/=k;tr=r;tr/=k;d--;
               td=cal(d);
               if(tl==tr)
               {
    
                   for(int i=l;i<=r;i++)
                   if(num[i]/td%10==p)
                   {
                       ans++;
                   }
                   printf("%d
    ",ans);
               }
               else
               {
                   for(int i=tl+1;i<tr;i++)
                   {
                       ans+=block[i].nt[d][p];
                   }
                   tl=(tl+1)*k;
                   for(int i=l;i<tl;i++)
                   if(num[i]/td%10==p)
                   {
                       ans++;
                   }
                   tr*=k;
                   for(int i=tr;i<=r;i++)
                   if(num[i]/td%10==p)
                   {
                       ans++;
                   }
                   printf("%d
    ",ans);
               }
               //cout<<"??"<<endl;
            }
        }
    }
    int main()
    {
        int cas,m,n;
        scanf("%d",&cas);
        while(cas--)
        {
            scanf("%d%d",&n,&m);
            int k=init(n);
            work(k,n,m);
        }
        return 0;
    }
    


    以下还写一写离线处理的代码,随后跟上。







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  • 原文地址:https://www.cnblogs.com/blfshiye/p/4278593.html
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