Argestes and Sequence

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

15 710 11 12 13 14Q 1 5 2 1Q 1 5 1 0Q 1 5 1 1Q 1 5 3 0Q 1 5 3 1S 1 100Q 1 5 3 1

Sample Output

511501
 
分块算法：大致就是把一堆(包含n个基本单位)东西分成sqrt(n)块， 每块包含sqrt(n)个基本单位。
        如果要修改某个基本单位，只要修改该单位所在的块，因为每块包含sqrt(n)个基本单位，所以时间复杂度为sqrt(n)；
      
view code#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long ll;const int N = 100010;struct Block{    int ct[10][10];}block[400];int _, n, m, a[N], cnt, pp[11];void build(){    int tmp = (int)sqrt(n*1.0), id;    cnt = n/tmp + 1;    memset(block, 0, sizeof(block));    for(int i=0; i<n; i++)    {        scanf("%d", &a[i]);        id = i/cnt, tmp = a[i];        for(int j=0; j<10; j++)        {            block[id].ct[j][tmp%10]++;            tmp /= 10;        }    }}void update(int u, int v){    int id = u/cnt;    for(int i=0; i<10; i++)    {        block[id].ct[i][a[u]%10]--;        a[u] /= 10;    }    a[u] = v;    for(int i=0; i<10; i++)    {        block[id].ct[i][v%10]++;        v /= 10;    }}int query(int l, int r, int d, int p){    int L = l/cnt, R = r/cnt;    int res = 0, div = pp[d];    if(L==R)    {        for(int i=l; i<=r; i++) if(a[i]/div%10==p) res++;        return res;    }    for(int i=L+1; i<R; i++)    {        res += block[i].ct[d][p];    }    for(int i=l; i<(L+1)*cnt; i++)    {        if(a[i]/div%10==p) res++;    }    for(int i=R*cnt; i<=r; i++)    {        if(a[i]/div%10==p) res++;    }    return res;}void solve(){    scanf("%d%d", &n, &m);    build();    char str[5];    int u, v, d, p;    while(m--)    {        scanf("%s", str);        if(str[0]==‘S‘)        {            scanf("%d%d", &u, &v);            u--;            update(u, v);        }        else        {            scanf("%d%d%d%d", &u, &v, &d, &p);            u--, v--, d--;            printf("%d\n", query(u, v, d, p));        }    }}void init(){    pp[0] = 1;    for(int i=1; i<10; i++) pp[i] = pp[i-1]*10;}int main(){//    freopen("in.txt", "r", stdin);    init();    cin>>_;    while(_--) solve();    return 0;}

 

hdu 5057 Argestes and Sequence(分块算法)