bzoj 4397: [Usaco2015 dec]Breed Counting -- 前缀和

4397: [Usaco2015 dec]Breed Counting

Time Limit: 10 Sec  Memory Limit: 128 MB

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，每个位置上的数只可能是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

#include<map>
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
char xB[1<<15],*xS=xB,*xTT=xB;
#define getc() (xS==xTT&&(xTT=(xS=xB)+fread(xB,1,1<<15,stdin),xS==xTT)?0:*xS++)
#define isd(c) (c>='0'&&c<='9')
inline int read(){
    char xchh;
    int xaa;
    while(xchh=getc(),!isd(xchh));(xaa=xchh-'0');
    while(xchh=getc(),isd(xchh))xaa=xaa*10+xchh-'0';return xaa;
}
int n,q,x,y,s[N][4];
int main()
{
    n=read();q=read();
    for(int i=1;i<=n;i++)
    {
        x=read();
        s[i][1]=s[i-1][1];
        s[i][2]=s[i-1][2];
        s[i][3]=s[i-1][3];
        s[i][x]++;
    }
    while(q--)
    {
        x=read();y=read();
        printf("%d %d %d
",s[y][1]-s[x-1][1],s[y][2]-s[x-1][2],s[y][3]-s[x-1][3]);
    }
    return 0;
}