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 BestCoder Round #11 (Div. 2)
题解：
这道题有三种版本的 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，其实还蛮简单的，就是将n组元素分成m组，每组合并成一块，查询时，只要看元素在那几块，相加就行了。
#include
#include
#include
#include
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\n",ans); } else { for(int i=tl+1;i
下面还写一写离线处理的代码，随后跟上。