bzoj4397[Usaco2015 dec]Breed Counting 前缀和

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

题解：用三个数组a,b,c来记录前i个字符里1,2,3的个数，运用前缀和的思想，区间l，r内的1,2,3,的个数即为a[r]-a[l-1]。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define maxn 100010using namespace std;int n,q,a[maxn],b[maxn],c[maxn];int main(){    scanf("%d%d",&n,&q);    for(int i=1;i<=n;i++)    {        int x;        scanf("%d",&x);        a[i]=a[i-1],b[i]=b[i-1],c[i]=c[i-1];        if(x==1)        a[i]++;        if(x==2)        b[i]++;        if(x==3)        c[i]++;    }    while(q--)    {        int x,y;        scanf("%d%d",&x,&y);        printf("%d %d %d\n",a[y]-a[x-1],b[y]-b[x-1],c[y]-c[x-1]);    }}