BZOJ4397Breed Counting

4397: [Usaco2015 dec]Breed Counting
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 77 Solved: 61
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
Source
Silver鸣谢Claris提供译文
直接求前缀和即可。。
1A。。
附上本蒟蒻的代码：

#include<cstdio>using namespace std;int n,m,a[4][100001];int read(){    int w=0,c=1; char ch=getchar();    while (ch<'0' || ch>'9')      {        if (ch=='-') c=-1;        ch=getchar();      }    while (ch>='0' && ch<='9')      w=w*10+ch-'0',ch=getchar();    return w*c;}int query(int x,int y,int z){    return a[x][z]-a[x][y-1];}int main(){    int i,x,y;    n=read(),m=read();    for (i=1;i<=n;i++)      {        x=read();        a[1][i]=a[1][i-1]+(x==1);        a[2][i]=a[2][i-1]+(x==2);        a[3][i]=a[3][i-1]+(x==3);      }    for (i=1;i<=m;i++)      {        x=read(),y=read();        printf("%d %d %d\n",query(1,x,y),query(2,x,y),query(3,x,y));      }    return 0;}