【loj 6277】 区间修改 单点查询

loj 6277
复杂度还是设为sqrt(n)
区间加就相当于暴力修改 和中间一段延迟数组区间加
如果x y 本身就在一个块内 那么就不需要加第二次

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<#define dbg3(x1,x2,x3) cout<<#x1<<" = "<#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 50025;
int pos[MAX_N],col[MAX_N],arr[MAX_N],block;
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,opt,x,y,v;scanf("%d",&n);
    block = sqrt(n);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]),pos[i] = (i-1)/block+1;
    for(int i = 1;i<=n;++i)
    {
        scanf("%d%d%d%d",&opt,&x,&y,&v);
        if(opt==1)
        {
            printf("%d\n",arr[y]+col[pos[y]]);
        }
        else
        {
            int st = pos[x]+1,ed = pos[y]-1;
            for(int j = st;j<=ed;j++)
                col[j] += v;
            for(int j = x;j<=min(pos[x]*block,y);j++)
                arr[j]+=v;
            if(pos[x]!=pos[y])
                for(int j = (pos[y]-1)*block+1;j<=y;++j)
                    arr[j]+=v;
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}