模板--------线段树的单点修改，区间查询

以区间求和为例

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define re register
#define inl inline
#define LL long long
const int MAXN=5e5+5;
LL a[MAXN],sum[4*MAXN];//一定要乘4
LL n,m,ans;
inl void BUD(LL num,LL l,LL r)
{
	if(l==r)
	{
		sum[num]=a[l];
		return;
	}
	LL mid=(l+r)/2;
	BUD(num*2,l,mid);
	BUD(num*2+1,mid+1,r);
	sum[num]=sum[num*2]+sum[num*2+1];
}
inl void CG(LL num,LL l,LL r,LL x,LL t)
{
	if(x<l||x>r) return;
	if(l==r&&l==x)
	{
		sum[num]+=t;
		return;
	}
	LL mid=(l+r)>>1;
	CG(num*2,l,mid,x,t);
	CG(num*2+1,mid+1,r,x,t);
	sum[num]=sum[num*2]+sum[num*2+1];
}
inl LL SUM(LL num,LL l,LL r,LL x,LL y)
{
	LL mid,res;
	if(y<l||x>r) return 0;
	if(x<=l&&y>=r) return sum[num];
	mid=(l+r)>>1;
	res=SUM(num*2,l,mid,x,y)+SUM(num*2+1,mid+1,r,x,y);
	return res;
}
int main() 
{
	LL ty,x,y;
	scanf("%lld%lld",&n,&m);
	for(re LL i=1;i<=n;i++)
	  scanf("%lld",&a[i]);
	BUD(1,1,n);
	while(m--)
	{
		scanf("%lld%lld%lld",&ty,&x,&y);
		if(ty==1) CG(1,1,n,x,y);
		else printf("%lld\n",SUM(1,1,n,x,y));
	}
	return 0;
}