程序師世界是廣大編程愛好者互助、分享、學習的平台,程序師世界有你更精彩!
首頁
編程語言
C語言|JAVA編程
Python編程
網頁編程
ASP編程|PHP編程
JSP編程
數據庫知識
MYSQL數據庫|SqlServer數據庫
Oracle數據庫|DB2數據庫
 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> bzoj4397【Usaco2015 Dec】Breed Counting

bzoj4397【Usaco2015 Dec】Breed Counting

編輯:C++入門知識

bzoj4397【Usaco2015 Dec】Breed Counting


 

4397: [Usaco2015 dec]Breed Counting

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.

 

給定一個長度為N的序列,每個位置上的數只可能是1,2,3中的一種。

有Q次詢問,每次給定兩個數a,b,請分別輸出區間[a,b]裡數字1,2,3的個數。

 

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

#include #include
#include
#include
#include
#include
#include
#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
#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]);
}
}

 

 

 

  1. 上一頁:
  2. 下一頁:
Copyright © 程式師世界 All Rights Reserved