洛谷P4145 上帝造题的⑦minutes ②

又是线段树。

区间开平方求和，套路题。

如果开到了1就不用再开下去了，否则直接到底。

记得 l > r 时交换 l r

 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 typedef long long LL;
 const int N = ;
 struct SegmentTree {
     LL sum[N << ];
     bool vis[N << ];

     inline void pushup(int l, int r, int o) {
         if(l == r) {
             if(sum[o] == ) {
                 vis[o] = ;
             }
             return;
         }
         sum[o] = sum[o << ] + sum[o <<  | ];
         vis[o] = (sum[o] == (r - l + ));
         return;
     }
     void change(int L, int R, int l ,int r, int o) {
         if(vis[o]) {
             return;
         }
         int mid = (l + r) >> ;
         if(l == r) {
             sum[o] = (LL)(sqrt((long double)(sum[o])));
             pushup(l, r, o);
             return;
         }
         if(L <= mid) {
             change(L, R, l, mid, o << );
         }
         if(mid < R) {
             change(L, R, mid + , r, o <<  | );
         }
         pushup(l, r, o);
         return;
     }
     LL ask(int L, int R, int l, int r, int o) {
         if(L <= l && r <= R) {
             return sum[o];
         }
         if(r < L || R < l) {
             return ;
         }
         int mid = (l + r) >> ;
         return ask(L, R, l, mid, o << ) + ask(L, R, mid + , r, o <<  | );
     }
     void build(int l, int r, int o) {
         if(l == r) {
             scanf("%lld", &sum[o]);
             pushup(l, r, o);
             return;
         }
         int mid = (l + r) >> ;
         build(l, mid, o << );
         build(mid + , r, o <<  | );
         pushup(l, r, o);
         return;
     }
 }SgT;

 int main() {
     int n;
     scanf("%d", &n);
     SgT.build(, n, );
     int m;
     scanf("%d", &m);
     int f, x, y;
     while(m--) {
         scanf("%d%d%d", &f, &x, &y);
         if(x > y) {
             std::swap(x, y);
         }
         if(f) {
             printf("%lld\n",SgT.ask(x, y, , n, ));
         }
         else {
             SgT.change(x, y, , n, );
         }
     }
     return ;
 }

AC代码