[HDU 4267](长春1001) 正式可以手写线段树了

跟孟神混，向着金牌！

方法一： 开55棵线段树

方法二： 线段树55个Node。

Attation：

1、线段树传Struct TLE

2、开55*50000*2 MLE

3、多点更新单点求值不用PushUp

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
const int N = 50001;
int zhuzhu[20][20];
struct Meng{
    int v[55],add;
    void clear(){
        memset(v,0,sizeof(v));
        add=0;
    }
    Meng operator + (const Meng & A) const{
        Meng ret;
        ret.add=0;
        for (int i = 0 ; i < 55 ; ++ i ){
            ret.v[i] = v[i] + A.v[i];
            ret.add += v[i] + A.v[i];
        }
        return ret;
    }
    void output(){
        for (int i = 0 ; i < 55 ; ++ i)
            printf(" %d" , v[i]);
        printf("\n");
    }
}tree[ N << 2 ];
int add[ N << 2 ];
void PushUp(int rt){
    tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}
void PushDown(int rt){
    if (add[rt]){
        tree[rt << 1] = tree[rt << 1] + tree[rt];
        tree[rt << 1 | 1] = tree[rt << 1 | 1] + tree[rt];
        add[rt << 1] = add[rt << 1] + add[rt];
        add[rt << 1 | 1] = add[rt << 1 | 1] + add[rt];
        add[rt]=0;
        tree[rt].clear();
    }
}
void build(int l , int r , int rt ){
    add[rt]=0;
    if (l == r){
        tree[rt].clear();
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}
void update(int L , int R , int k , int c , int l, int r, int rt){
    if (L <= l && r <= R){
        tree[rt].v[ zhuzhu[k][L%k] ] += c;
        tree[rt].add += c;
        add[rt]+= c;
        return;
    }
    PushDown(rt);
    int m = (l + r) >> 1;
    if (L <= m) update(L , R , k , c , lson);
    if (m < R) update(L , R , k , c , rson);
    //PushUp(rt);
}
int query(int p , int l , int r ,int rt){
    if (l == r) {
        int zhu = 0 , res = 0;
        for (int i = 1; i <= 10 ; ++ i)
            res += tree[rt].v[ zhuzhu[i][p%i] ];
        return res;
    }
    PushDown(rt);
    int m = (l + r) >> 1;
    if (p <= m) return query(p , lson);
    return query(p , rson);
}
int a[N] , n;
void solve(){
    for (int i = 1 ; i <= n ; ++ i){
        scanf("%d",&a[i]);
    }
    build(1 , n , 1);
    int m;
    scanf("%d", &m);
    while (m--){
        int op;
        scanf("%d", &op);
        if (op==2){
            int x;
            scanf("%d", &x);
            //res.output();
            int ans = a[x] + query(x,1,n,1);
            printf("%d\n",ans);
        }
        else{
            int a,b,k,c;
            scanf("%d%d%d%d",&a,&b,&k,&c);
            update(a,b,k,c,1,n,1);
        }
    }
}
void init(){
    int zhu=0;
    for (int i = 1 ; i <= 10 ; ++ i)
        for (int j = 0 ; j < i ;  ++j ){
            zhuzhu[i][j]=zhu++;
        }
}
int main(){
    init();
    while (~scanf("%d",&n)) solve();
}