Argestes and Sequence
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1279 Accepted Submission(s): 373
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.
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.
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]<=231 - 1
1<=X<=N
0<=Y<=231 - 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
1
分块大法，降低暴力的时间复杂度
//2016.8.13
#include<iostream>
#include<cstdio>
#include<cstring> using namespace std; const int N = ;
int a[N], len = , n, mi[] = {, , , , , , , , , , }; struct node
{
int cnt[][];//cnt[d][p]表示每个块内第d位是p的个数
}block[]; bool judge(int x, int d, int p)//判断x的第d位是否为p
{
return x/mi[d]%==p;
} int query(int l, int r, int d, int p)
{
int id1 = l/len, id2 = r/len, ans = ;
if(id1==id2)//如果l,r在同一个块内，暴力枚举
{
for(int i = l; i <= r; i++)
if(judge(a[i], d, p))
ans++;
}else
{
for(int i = id1+; i <= id2-; i++)//把中间的块加起来
ans+=block[i].cnt[d][p];
for(int i = l; i <= (id1+)*len && i <= n; i++)//暴力最左边的块
if(judge(a[i], d, p))
ans++;
for(int i = id2*len+; i <= r; i++)//暴力最右边的块
if(judge(a[i], d, p))
ans++;
}
return ans;
} void update(int x, int y)
{
int tmp = a[x], id = (x-)/len;//这里起初x少减了个1，WA了好多次忧伤。。。
a[x] = y;
for(int i = ; i <= ; i++)
{
block[id].cnt[i][tmp%]--;
tmp/=;
}
tmp = a[x];
for(int i = ; i <= ; i++)
{
block[id].cnt[i][tmp%]++;
tmp/=;
}
} int main()
{
int T, d, p, m, x, y, l, r, id;
char cmd;
cin>>T;
while(T--)
{
scanf("%d%d", &n, &m);
memset(block, , sizeof(block));
memset(a, , sizeof(a));
for(int i = ; i <= n; i++)
{
scanf("%d", &a[i]);
id = (i-)/len;
int tmp = a[i];
for(int j = ; j <= ; j++)//初始化块
{
block[id].cnt[j][tmp%]++;
tmp/=;
}
}
while(m--)
{
getchar();
scanf("%c", &cmd);
if(cmd == 'Q')
{
int ans = ;
scanf("%d%d%d%d", &l, &r, &d, &p);
ans = query(l, r, d, p);
printf("%d\n", ans);
}else
{
scanf("%d%d", &x, &y);
update(x, y);
}
}
} return ;
}