[BZOJ3211]花神游历各国 势能线段树

势能线段树
维护区间不为1的元素数量
对于存在元素不为1的区间暴力递归修改

/**************************************************************
    Problem: 3211
    User: di4CoveRy
    Language: C++
    Result: Accepted
    Time:2164 ms
    Memory:7928 kb
****************************************************************/

#include 
#include 
#include 
#define N 100050
#define COUT "%lld\n"

using namespace std;

long long ans;

struct Tree{long long sum,b;}tree[N*4];
int d[N],n,m;
Tree merge(Tree p1,Tree p2)
{
    Tree tmp;
    tmp.sum = p1.sum + p2.sum;
    tmp.b = p1.b + p2.b;
    return tmp;
}

void build(int l,int r,int t)
{
    if (l == r)
    {
        tree[t].sum = 1LL * d[l];
        tree[t].b = 1;
        return ;
    }
    int mid = (l + r) / 2;
    build(l,mid,2*t);
    build(mid+1,r,2*t+1);
    tree[t] = merge(tree[2*t],tree[2*t+1]);
}

int ll,rr;
void update(int l,int r,int t)
{
    if (l > rr || r < ll) return ;
    if (l >= ll && r <= rr && tree[t].b == 0) return ;
    if (l >= ll && r <= rr && l == r)
    {
        tree[t].sum = 1LL * sqrt(tree[t].sum);
        if (tree[t].sum <= 1) tree[t].b = 0;else tree[t].b = 1;
        return ; 
    }
    int mid = (l + r) / 2;
    update(l,mid,2*t);
    update(mid+1,r,2*t+1);
    tree[t] = merge(tree[2*t],tree[2*t+1]);
}

void query(int l,int r,int t)
{
    if (l > rr || r < ll) return ;
    if (l >= ll && r <= rr) {ans += tree[t].sum;  return ;}
    int mid = (l+r) / 2;
    query(l,mid,2*t);
    query(mid+1,r,2*t+1);
}

int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%d",&d[i]);
    build(1,n,1);
    scanf("%d",&m);
    for (int i=1;i<=m;i++)
    {
        int cmd = 0;
        scanf("%d%d%d",&cmd,&ll,&rr);
        if (cmd == 2) update(1,n,1);
        if (cmd == 1) 
        {
            ans = 0LL;
            query(1,n,1);
            printf(COUT,ans);
        }
    }
    return 0;
}