bzoj4397【Usaco2015 Dec】Breed Counting

4397: [Usaco2015 dec]Breed Counting

Time Limit: 10 Sec  Memory Limit: 128 MB

Submit: 29  Solved: 25

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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 would like your help counting the number of cows of each breed that lie within certain intervals of the ordering.

给定一个长度为N的序列，每一个位置上的数仅仅可能是1，2，3中的一种。

有Q次询问，每次给定两个数a,b。请分别输出区间[a,b]里数字1，2。3的个数。

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 a…b that are Holsteins (breed 1), Guernseys (breed 2), and Jerseys (breed 3).

Sample Input

6 3

2

1

1

3

2

1

1 6

3 3

2 4

Sample Output

3 2 1

1 0 0

2 0 1

HINT

Source

search=Silver%E9%B8%A3%E8%B0%A2Claris%E6%8F%90%E4%BE%9B%E8%AF%91%E6%96%87" style="color:blue; text-decoration:none">Silver鸣谢Claris提供译文

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
#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 long
#define pa pair<int,int>
#define maxn 100005
#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 %d %d\n",sum[1][y]-sum[1][x-1],sum[2][y]-sum[2][x-1],sum[3][y]-sum[3][x-1]);
	}
}