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
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#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
以下还写一写离线处理的代码，随后跟上。