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.
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
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\n",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\n",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。