Bzoj4397 [usaco 2015 Dec] Breed Counting

Bzoj4397 [usaco 2015 Dec] Breed Counting

 

4397: [usaco dec] Breed Counting Description

Farmer John's N cows, conveniently numbered 1... N, are all standing in a row (they seem to do so often that it now takes very little prompting from Farmer John to line them up ). each cow has a breed ID: 1 for Holsteins, 2 for Guernseys, and 3 for Jerseys. farmer John wowould like your help counting the number of cows of each breed that lie within certain intervals of the ordering.

 

Given a sequence with a length of N, the number at each position can only be one of 2, 3.

For Q-related queries, each time two numbers a and B are given, please output the numbers 1, 2, and 3 in the range [a, B] respectively.

 

Input

The first line of input contains NN and QQ (1 ≤ N ≤ 100,000 1 ≤ Q ≤ 100,000 ).
The next NN lines contain an integer that is either 1, 2, or 3, giving the breed ID of a single cow in the ordering.
The next QQ lines describe a query in the form of two integers a, B (a ≤ B ).

Output

For each of the QQ queries (a, B), print a line containing three numbers: the number of cows numbered... B that are Holsteins (breed 1), Guernseys (breed 2), and Jerseys (breed 3 ).

Sample Input6 3
2
1
1
3
2
1
1 6
3 3
2 4 Sample Output3 2 1
1 0 0
2 0 1 HINT

 

Source # include # Include
# Include
# Include
# Include
# Include
# Include
# Define F (I, j, n) for (int I = j; I <= n; I ++)
# Define D (I, j, n) for (int I = j; I> = n; I --)
# Define ll long
# Define pa pair
# Define maxn100005
# Define inf 1000000000
Using namespace std;
Int n, m, x, y, sum [4] [maxn];
Inline int read ()
{
Int x = 0, f = 1; char ch = getchar ();
While (ch <'0' | ch> '9') {if (ch = '-') f =-1; ch = getchar ();}
While (ch> = '0' & ch <= '9') {x = x * 10 + ch-'0'; ch = getchar ();}
Return x * f;
}
Int main ()
{
N = read (); m = read ();
F (I, 1, n)
{
F (j, 1, 3) sum [j] [I] = sum [j] [I-1];
X = read ();
Sum [x] [I] ++;
}
F (I, 1, m)
{
X = read (); y = read ();
Printf ("% d \ n", sum [1] [y]-sum [1] [x-1], sum [2] [y]-sum [2] [x-1], sum [3] [y]-sum [3] [x-1]);
}
}