HDU 6315 Naive Operations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 502768/502768 K (Java/Others)
Problem Description
In a galaxy far, far away, there are two integer sequence $a$ and $b$ of length $n$.
$b$ is a static permutation of $1$ to $n$. Initially $a$ is filled with zeroes.
There are two kind of operations:
1. add l r: add one for $a_l$，$a_{l+1}$……$a_r$
2. query l r: query $\sum_{i=l}^{r}\biggl\lfloor\frac{a_i}{b_i}\biggr\rfloor$
Input
There are multiple test cases, please read till the end of input file.
For each test case, in the first line, two integers n,q, representing the length of a,b and the number of queries.
In the second line, n integers separated by spaces, representing permutation b.
In the following q lines, each line is either in the form 'add l r' or 'query l r', representing an operation.
1≤n,q≤100000, 1≤l≤r≤n, there're no more than 5 test cases.
Output
Output the answer for each 'query', each one line.
Sample Input
5 12
1 5 2 4 3
add 1 4
query 1 4
add 2 5
query 2 5
add 3 5
query 1 5
add 2 4
query 1 4
add 2 5
query 2 5
add 2 2
query 1 5
Sample Output
1
1
2
4
4
6
解题思路
这题可优化的地方在于a[i]要增加b[i]次才能对答案产生1的贡献。线段树更新时，先将区间内maxa加一，然后判断，对于区间内a的最大值小于等于b的最小值的，我们只更新maxa，不更新这个区间对答案的贡献ans，然后lazy标记打上，返回。对于最大值大于等于b的最小值的，我们接着更新下去，直到叶子结点，这样不会超时，原因先留坑
填坑啦2019年03月09日15:48:41
https://blog.csdn.net/weixin_38287798/article/details/81210396 关键就在于b数组是1到n的全排列，不能全是1，所以可以暴力到叶子
源代码
1 #include<stdio.h> 2 #include<algorithm> 3 4 int n,q; 5 int b[100010]; 6 7 struct Segtree{ 8 int l,r,ans; 9 int maxa,minb; 10 }t[400010]; 11 int lazy[400010]; 12 inline int lson(int x){return x<<1;} 13 inline int rson(int x){return x<<1|1;} 14 void maketree(int x,int l,int r) 15 { 16 lazy[x]=0; 17 if(l==r) 18 { 19 t[x]={l,r,0,0,b[l]}; 20 return; 21 } 22 t[x]={l,r,0,0,0}; 23 int mid=(l+r)>>1; 24 maketree(lson(x),l,mid); 25 maketree(rson(x),mid+1,r); 26 t[x].minb=std::min(t[lson(x)].minb,t[rson(x)].minb); 27 } 28 void pushdown(int x) 29 { 30 if(!lazy[x]) return; 31 t[lson(x)].maxa+=lazy[x]; 32 t[rson(x)].maxa+=lazy[x]; 33 lazy[lson(x)]+=lazy[x]; 34 lazy[rson(x)]+=lazy[x]; 35 lazy[x]=0; 36 } 37 void pushup(int x) 38 { 39 if(t[x].l==t[x].r) return; 40 t[x].minb=std::min(t[lson(x)].minb,t[rson(x)].minb); 41 t[x].maxa=std::max(t[lson(x)].maxa,t[rson(x)].maxa); 42 t[x].ans=t[lson(x)].ans+t[rson(x)].ans; 43 } 44 void update(int x,int l,int r)//l到r的a增加1 45 { 46 if(l>t[x].r||r<t[x].l) return; 47 if(l<=t[x].l&&t[x].r<=r) 48 { 49 t[x].maxa++; 50 if(t[x].maxa<t[x].minb) 51 { 52 lazy[x]++; 53 return; 54 } 55 if(t[x].l==t[x].r&&t[x].maxa>=t[x].minb) 56 { 57 t[x].ans++; 58 t[x].minb+=b[t[x].l];//为下次更新t[x].ans做准备 59 return; 60 } 61 } 62 pushdown(x); 63 update(lson(x),l,r); 64 update(rson(x),l,r); 65 pushup(x); 66 } 67 int que(int x,int l,int r) 68 { 69 if(l>t[x].r||r<t[x].l) return 0; 70 if(l<=t[x].l&&t[x].r<=r) 71 return t[x].ans; 72 pushdown(x); 73 return que(lson(x),l,r)+que(rson(x),l,r); 74 } 75 76 int main() 77 { 78 //freopen("test.in","r",stdin); 79 while(~scanf("%d%d",&n,&q)) 80 { 81 for(int i=1;i<=n;i++) scanf("%d",b+i); 82 maketree(1,1,n); 83 while(q--) 84 { 85 char s[10]; 86 int l,r; 87 scanf("\n%s %d %d",s,&l,&r); 88 if(s[0]=='a') update(1,l,r); 89 else printf("%d\n",que(1,l,r)); 90 } 91 } 92 return 0; 93 }
