阎帝小代码

#include<bits/stdc++.h>
#define int long long
#define lc(p) (p<<1)
#define rc(p) (p<<1|1)
using namespace std;
const int N=6e5+10;//不清楚为什么要开6e5,我开5e5就RE 
struct node{int l,r,sum;}t[N<<2];
int a[N],n,q;
void pu(int p){t[p].sum=t[lc(p)].sum+t[rc(p)].sum;}
void build(int id,int l,int r)
{
	t[id]={l,r,0};
	if(l==r){t[id].sum=a[l];return ;}
	int m=l+r>>1;
	build(lc(id),l,m);build(rc(id),m+1,r);
	pu(id);
}
void change(int p,int x,int y)
{
	if(x<t[p].l||x>t[p].r)return ;
	t[p].sum+=y;
	change(lc(p),x,y);change(rc(p),x,y);
}
int query(int p,int l,int r)
{
	if(r<t[p].l||t[p].r<l)return 0;
	if(l<=t[p].l&&t[p].r<=r)return t[p].sum;
	return query(lc(p),l,r)+query(rc(p), l,r);
}
signed main()
{
	scanf("%lld%lld",&n,&q);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
	build(1,1,n);
	while(q--)
	{
		int op,x,y;scanf("%lld%lld%lld",&op,&x,&y);x++;
		if(op==0)
		{
			change(1,x,y);
		}
		else
		{
			printf("     %lld\n",query(1,x,y));
		}
	}
	return 0;
}

点修区查，这不树状数组板子吗。

注意下标需要整体加 $1$，树状数组本身不支持下标 $0$ 作为有效操作节点。

提交记录：626ms，8 MiB这不比线段树香吗

#include<bits/stdc++.h>
using namespace std;
#define int ll
#define In inline
#define ll long long 
#define rep(i,x,y) for(int i=x;i<=y;++i)
#define per(i,x,y) for(int i=x;i>=y;--i)
In void ios_off(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);}
const int N=5e5+10;
int n,q,c[N];
int lowbit(int x){return x&-x;}
void add(int x,int k){for(;x<=n;x+=lowbit(x))c[x]+=k;}
int query(int x)
{
	int res=0;
	for(;x;x-=lowbit(x))res+=c[x];
	return res;
}
signed main()
{
	ios_off();
	cin>>n>>q;
	rep(i,1,n)
	{
		int x;cin>>x;
		add(i,x);
	}
	rep(i,1,q)
	{
		int op;cin>>op;
		if(op==0)
		{
			int p,x;cin>>p>>x;
			add(p+1,x);
		}
		else
		{
			int l,r;cin>>l>>r;
			cout<<query(r)-query(l)<<'\n';
		}
	}
	return 0;
}

你是说不想在以后出现函数名重复的情况？

那就试试结构体吧！！！

线段树：

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=5e5+10;
int a[N];
#define lc(p) (p<<1)
#define rc(p) (p<<1|1)
struct node{int l,r,s;};
struct SMT{
	node tr[N<<2];
	void pushup(int p){tr[p].s=tr[lc(p)].s+tr[rc(p)].s;}
	void bt(int p,int l,int r)
	{
		tr[p]={l,r,0};
		if(l==r){tr[p].s=a[l];return;}
		int mid=(l+r)>>1;
		bt(lc(p),l,mid);bt(rc(p),mid+1,r);
		pushup(p);
	}
	void change(int p,int x,int k)
	{
		if(tr[p].l>x||tr[p].r<x)return ;
		if(tr[p].l==tr[p].r)
		{
			tr[p].s+=k;
			return;
		}
		change(lc(p),x,k);change(rc(p),x,k);
		pushup(p);
	}
	int query(int p,int l,int r)
	{
		if(tr[p].l>r||tr[p].r<l)return 0;
		if(l<=tr[p].l&&tr[p].r<=r)return tr[p].s;
		return query(lc(p),l,r)+query(rc(p),l,r);
	}
}tr;
#undef lc(p)
#undef rc(p)
signed main()
{
	int n,q;cin>>n>>q;
	for(int i=1;i<=n;i++)cin>>a[i];
	tr.bt(1,1,n);
	while(q--)
	{
		int op,x,y;cin>>op>>x>>y;
		if(op==0)tr.change(1,x+1,y);
		else cout<<tr.query(1,x+1,y)<<'\n';
	}
	return 0;
}

树状数组：

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=5e5+10;
struct BIT{
	int c[N],n;
	void add(int x,int k){for(;x<=n;x+=x&-x)c[x]+=k;}
	int ask(int x){int ans=0;for(;x;x-=x&-x)ans+=c[x];return ans;}
}tr;
int a[N];
signed main()
{
	int n,q;cin>>n>>q;tr.n=n;
	for(int i=1;i<=n;i++)cin>>a[i],tr.add(i,a[i]);
	while(q--)
	{
		int op,x,y;cin>>op>>x>>y;
		if(op==0)tr.add(x+1,y);
		else cout<<(tr.ask(y)-tr.ask(x))<<'\n';
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define lc(p) (p << 1)      /*i的左孩子编号为i*2*/
#define rc(p) (p << 1 | 1)  /*i的右孩子编号为i*2+1*/
const int N = 6e5 + 10;
#define int long long
int a[N];/*记录数组初值*/
struct node
{int l/*左区间*/, r/*右区间*/, v/*区间最小值*/;}tr[N << 2]/*开四倍, 防炸*/;
void pushup(int p){tr[p].v = tr[lc(p)].v + tr[rc(p)].v;}/*更新节点权值*/
void build(int p, int l, int r)/*节点p管理区间[l,r]构造线段树*/
{
    tr[p] = {l, r, 0};/*赋值节点p*/
if (l == r)/*如果只管理一个点*/
        {tr[p].v = a[l];/*直接赋值权值*/ return;/*已经到最低层了*/ }
    int mid = (l + r) >> 1;
    build(lc(p), l, mid);/*左孩子管理p管理范围的左半边*/
    build(rc(p), mid + 1, r);/*右孩子管理剩下右半边*/
    pushup(p);/*递归回来时更新p的区间和*/
}
int query(int p, int l, int r)/*查询区间[l,r]的区间和*/
{
    if (r < tr[p].l or tr[p].r < l)/*如果p管理区间与[l,r]无关*/
        return 0;/*对答案贡献为0*/
    if (l <= tr[p].l and tr[p].r <= r)/*p区间在区间[l,r]内*/
        return tr[p].v;/*不会贡献区间以外的答案, 直接返回p的权值*/

    /*如果p包括了查询区间以外的值, 就将区间细分到左右儿子再统计*/
    return query(lc(p), l, r) + query(rc(p), l, r);
}
void change(int p, int x, int y)/*将a[x]的值加上y*/
{
    if (x < tr[p].l or tr[p].r < x)/*如果x不在p管理区间内*/
        return;
    tr[p].v += y;/*如果x在p管理区间内, p的区间和加y*/
    change(lc(p), x, y); change(rc(p), x, y);/*递归修改左右孩子的区间和*/
}
signed main()
{
    int n, q; cin >> n >> q;
    for (int i = 1; i <= n; i++) cin >> a[i];
    build(1, 1, n);
    while (q--)
    {
        int op, x, y; cin >> op >> x >> y;
        x ++;
        if (op == 0)
            change(1, x, y);
        else
            cout << query(1, x, y) << '\n';
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define lc(p) tr[p].ls
#define rc(p) tr[p].rs
const int N = 5e5 + 10;
struct node{int val, ls, rs;}tr[N << 2];
int trlen = 1, rt, a[N];
void pushup(int p){tr[p].val = tr[lc(p)].val + tr[rc(p)].val;}

void build(int p, int l, int r)
{
    tr[p] = {0, 0, 0};
    if (l == r) {tr[p].val = a[l]; return;}
    int mid = (l + r) >> 1;
    lc(p) = ++trlen; build(lc(p), l, mid);
    rc(p) = ++trlen; build (rc(p), mid + 1, r);
    pushup(p);
}

void change(int p, int l, int r, int k, int x)
{
    if (l == r) {tr[p].val += x; return; }
    int mid = (l + r) >> 1;
    if (k <= mid) change(lc(p), l, mid, k, x);
    else change(rc(p), mid + 1, r, k, x);
    pushup(p);
}

int query(int p, int l, int r, int x, int y)
{
    if (y < l or r < x) return 0;
    if (x <= l and r <= y) return tr[p].val;
    int mid = (l + r) >> 1;
    return query(lc(p), l, mid, x, y) + query(rc(p), mid + 1, r, x, y);
}

signed main()
{
    int n, q; cin >> n >> q;
    for (int i = 1; i <= n; i++) cin >> a[i];
    build(1, 1, n);
    while (q--)
    {
        int op, x, y; cin >> op >> x >> y;
        if (op == 0) change(1, 1, n, x + 1, y);
        else if (op == 1) cout << query(1, 1, n, x + 1, y) << endl;
    }
    return 0;
}