NBUT 1602 Mod Three（线段树单点更新区间查询）

• [1602] Mod Three

• 时间限制: 5000 ms　内存限制: 65535 K
• 问题描述
• Please help me to solve this problem, if so, LiangChen must have zhongxie!
  There is a sequence contains n integers a0, a1, ...., an, firstly all of them equal 0,
  and there are q opertions, each of them either is
  

  0 x; add ax by 1

  or
  1 l r; calculator the total number of ai that ai mod 3 == 0 (l <= i <= r)

• 输入
• Input starts with an integer T denoting the number of test cases.
  For each test case, first line contains n and q(1 <= n, q <= 100,000), next line contain n integers.
• 输出
• For each test case, first line contains "Case X:", then print your answer, one line contains one integer.
• 样例输入

• 1
  10 8
  0 4
  0 3
  0 6
  1 2 3
  0 7
  1 1 10
  0 8
  1 1 8

• 样例输出

• Case 1:
  1
  6
  3

题目链接：NBUT 1602

看了一下居然跟是1603重复的，不过这题没人写，另外题目描述有点问题，序列不是a0~an而是a1~an。

用val统计叶子节点此时的值，cnt统计区间（节点）内模3不为0的个数，一开始都是0因此cnt的初始值为1

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=100010;
struct seg
{
	int l,mid,r;
	int val,cnt;
};
seg T[N<<2];
void pushup(int k)
{
	T[k].val=T[LC(k)].val+T[RC(k)].val;
	T[k].cnt=T[LC(k)].cnt+T[RC(k)].cnt;
}
void build(int k,int l,int r)
{
	T[k].l=l;
	T[k].r=r;
	T[k].mid=MID(l,r);
	if(l==r)
	{
		T[k].val=0;
		T[k].cnt=1;
		return ;
	}
	else
	{
		build(LC(k),l,T[k].mid);
		build(RC(k),T[k].mid+1,r);
		pushup(k);
	}
}
void update(int k,int x)
{
	if(T[k].l==T[k].r&&T[k].l==x)
		T[k].cnt=((++T[k].val)%3==0);
	else
	{
		if(x<=T[k].mid)
			update(LC(k),x);
		else
			update(RC(k),x);
		pushup(k);
	}
}
int query(int k,int l,int r)
{
	if(l<=T[k].l&&T[k].r<=r)
		return T[k].cnt;
	else
	{
		if(r<=T[k].mid)
			return query(LC(k),l,r);
		else if(l>T[k].mid)
			return query(RC(k),l,r);
		else
			return query(LC(k),l,T[k].mid)+query(RC(k),T[k].mid+1,r);
	}
}
int main(void)
{
	int tcase,i,n,m,ops,l,r,x;
	scanf("%d",&tcase);
	for (int q=1; q<=tcase; ++q)
	{
		scanf("%d%d",&n,&m);
		build(1,1,n);
		printf("Case %d:
",q);
		for (i=0; i<m; ++i)
		{
			scanf("%d",&ops);
			if(ops==0)
			{
				scanf("%d",&x);
				update(1,x);
			}
			else if(ops==1)
			{
				scanf("%d%d",&l,&r);
				printf("%d
",query(1,l,r));
			}
		}
	}
	return 0;
}