一维和二维树状数组

#include<cstdio>//注意0的死循环
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long int LL;
const LL mod=1e9+7;
LL n,m,x[1000010],arr[1010][1010];
LL lowbit(LL t)//求低位 0 会陷入死循环
{
    return t&(-t);
}

void update(LL i,LL add)//更新i节点加上add
{
    while(i<=n)
    {
        x[i]+=add;
        i+=lowbit(i);
    }
}

LL getans(LL s)//求前s项和
{
    LL sum=0;
    while(s>0)
    {
        sum+=x[s];
        s-=lowbit(s);
    }
    return sum;
}

void update2(LL a,LL b,LL add)//在x[a][b]处加add
{
    for(LL i=a; i<=n+1; i+=lowbit(i))
        for(LL j=b; j<=n+1; j+=lowbit(j))
            arr[i][j]+=add;
}
LL getans2(LL a,LL b)//求x[a][b]的前面之和
{
    LL sum=0;
    for(LL i=a; i>0; i-=lowbit(i))
        for(LL j=b; j>0; j-=lowbit(j))
            sum+=arr[i][j]%mod;
    return sum%mod;
}
LL getans3(LL x1,LL y1,LL x2,LL y2)//求x1,y1,  x2,y2的矩形里面之和
{
    LL ans=getans2(x2,y2)-getans2(x1-1,y2)-getans2(x2,y1-1)+getans2(x1-1,y1-1);
    while(ans<0) ans+=mod;
    return ans;
}
int main()
{
    scanf("%lld%lld",&n,&m);
    memset(x,0,sizeof(x));
    memset(arr,0,sizeof(arr));
    string str;
    while(m--)
    {
        cin>>str;
        LL a,b,c,d;
        if(str[0]=='S')
        {
            scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
            printf("%lld\n",getans3(a+1,b+1,c+1,d+1));
        }
        else
        {
            scanf("%lld%lld%lld",&a,&b,&c);
            update2(a+1,b+1,c);
        }
    }
}