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;
}