hdu5057 Argestes and Sequence 分块

Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)
Total Submission(s):     Accepted Submission(s): 

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[],a[],...,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[],a[],...,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]
<=T<=
<=N, M<=
<=a[i]<= -
<=X<=N
<=Y<= -
<=L<=R<=N
<=D<=
<=P<=

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

Sample Input

Q
Q
Q
Q
Q
S
Q    

Sample Output

Source
BestCoder Round # (Div. )

之前用线段树一直MLE(在码之前看了下内存限制，就觉得会开不下的了),知道可以用分块做之后，就看了下大白的一道分块题，再想这道题就比较简单了
~num[b][a][b]表示第b块所有的数在第a位为b的个数，这些都可以简单预处理出来，然后只是单点修改，把原来的值和新值对比一下更新一下就好了
#include <bits/stdc++.h>

using namespace std;
const int N = 1e5 + ;
const int SIZE = ;
int n, m;
int A[N];
int block[N / SIZE + ][SIZE + ];
int num[N / SIZE + ][][];
void pre(int b, int j)
{
    int *B = &block[b][];
    for(int i = ; i < j; ++i) {
        int tmp = B[i];
        int cnt = ;
        while(cnt <= ) {
            int x = tmp % ;
            tmp /= ;
            num[b][cnt][x]++;
            cnt++;
        }
    }
}
void init()
{
    memset(num, , sizeof num);
    scanf("%d%d", &n, &m);
    int j = , b = ;
    for(int i = ; i < n; ++i) {
        scanf("%d", &A[i]);
        block[b][j] = A[i];
        if(++j == SIZE) {
            pre(b, j);
            b++; j = ;
        }
    }
    if(j) { pre(b, j); ++b; }
}
int get(int x, int d) {
    int cnt = ;
    while(cnt < d) {
        x /= ;
        cnt++;
    }
    return x % ;
}
int query(int L, int R, int D, int p)
{
    int res = ;
    int lb = L / SIZE, rb = R / SIZE;
    if(lb == rb) {
        for(int i = L; i <= R; ++i) {
            if(get(A[i], D) == p) res++;
        }
    }
    else {
        for(int i = L; i < (lb + ) * SIZE; ++i) if(get(A[i], D) == p) res++;
        for(int i = rb * SIZE; i <= R; ++i) if(get(A[i], D) == p) res++;
        for(int i = lb + ; i < rb; ++i) res += num[i][D][p];
    }
    return res;
}
void modify(int x, int y)
{
    if(A[x] == y) return;
    int old = A[x], now = y, b = x / SIZE, cnt = ;
    int c1[], c2[];
    A[x] = y;
    while(cnt <= ) {
        c1[cnt] = old % ;
        c2[cnt] = now % ;
        old /= ;
        now /= ;
        cnt++;
    }
    for(int i = ; i <= ; ++i) {
        if(c1[i] != c2[i]) {
            num[b][i][ c2[i] ]++;
            num[b][i][ c1[i] ]--;
        }
    }
}
int main()
{
    int _; scanf("%d", &_);
    while(_ --)
    {
        init();
        char op[];
        int L, R, D, P;
        while(m --)
        {
            scanf("%s", op);
            if(op[] == 'Q') {
                scanf("%d%d%d%d", &L, &R, &D, &P);
                L--; R--;
                printf("%d\n", query(L, R, D, P));
            }else {
                scanf("%d%d", &D, &P);
                D--;
                modify(D, P);
            }
        }
    }
    return ;
}