题目链接：https://vjudge.net/contest/332656#problem/D
思路：因为根号运算n衰减的很快，所以在极少数的操作内它就会变成1，所以当整个区间内的值都变成1时直接返回，反之暴力更新叶子结点就好
1 #include <math.h> 2 #include <stdio.h> 3 #include <iostream> 4 #include <algorithm> 5 #include <string> 6 #include <string.h> 7 #include <vector> 8 #include <map> 9 #include <random> 10 11 12 #define LL long long 13 14 const int maxn = 1e5 + 10; 15 16 17 LL arr[maxn]; 18 struct segment_tree { 19 int l,r; 20 LL val; 21 }tree[maxn<<2]; 22 23 24 void pushup(int nod ) { 25 tree[nod].val = (tree[nod<<1].val + tree[(nod<<1)+1].val); 26 } 27 28 29 void build(int l,int r,int nod) { 30 tree[nod].l = l; 31 tree[nod].r = r; 32 if (l == r) { 33 tree[nod].val = arr[l]; 34 return ; 35 } 36 int mid = (l + r) >> 1; 37 build(l,mid,nod<<1); 38 build(mid+1,r,(nod<<1)+1); 39 pushup(nod); 40 } 41 42 void modify(int x,int y,int k=1) { 43 int l = tree[k].l,r = tree[k].r; 44 if (l == r) { 45 tree[k].val = std::sqrt(tree[k].val); 46 return ; 47 } 48 if (x <= l && y >= r && tree[k].val == r-l+1) 49 return ; 50 int mid = (l + r ) >> 1; 51 if (x <= mid) 52 modify(x,y,k<<1); 53 if (y > mid) { 54 modify(x,y,(k<<1)+1); 55 } 56 pushup(k); 57 } 58 59 LL query(int x,int y,int k=1) { 60 int l = tree[k].l,r = tree[k].r; 61 if (x <= l && y >= r) { 62 return tree[k].val; 63 } 64 int mid = (l + r) >> 1; 65 LL sum = 0; 66 if (x <= mid) 67 sum += query(x,y,k<<1); 68 if (y > mid) { 69 sum += query(x,y,(k<<1)+1); 70 } 71 return sum; 72 } 73 74 int main() { 75 int n; 76 int t = 1; 77 while (~scanf("%d",&n)) { 78 for (int i=1;i<=n;i++) { 79 scanf("%lld",&arr[i]); 80 } 81 build(1,n,1); 82 int m; 83 scanf("%d",&m); 84 printf("Case #%d: ",t++); 85 int x,y,z; 86 while (m--) { 87 scanf("%d%d%d",&z,&x,&y); 88 if (x > y) 89 std::swap(x,y); 90 if (!z) { 91 modify(x,y); 92 } 93 else { 94 printf("%lld ",query(x,y)); 95 } 96 } 97 printf(" "); 98 } 99 return 0; 100 }