Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 502768/502768 K (Java/Others) Total Submission(s): 1244 Accepted Submission(s): 519
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 al,al+1...ar 2. query l r: query ∑ri=l⌊ai/bi⌋
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
思路： 保存区间的sum，最大的a和(maxa)最小的b(minb), 若maxa<minb则没有必要向下更新 否则一直向下更新，一直到达要更新的点，然后tree[rt].sum++,tree[rt].minb+=b[l]。 代码： #include <iostream> #include <cstdio> #include <string.h> #include <algorithm> using namespace std; const int maxn=1e5+10; struct node { int sum,maxa,minb,add; }tree[maxn<<2]; int n,Q; int b[maxn]; void push_up(int rt) { tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum; tree[rt].maxa=max(tree[rt<<1].maxa,tree[rt<<1|1].maxa); tree[rt].minb=min(tree[rt<<1].minb,tree[rt<<1|1].minb); } void push_down(int rt) { if(tree[rt].add) { tree[rt<<1].add+=tree[rt].add; tree[rt<<1|1].add+=tree[rt].add; tree[rt<<1].maxa+=tree[rt].add; tree[rt<<1|1].maxa+=tree[rt].add; tree[rt].add=0; } } void build(int l,int r,int rt) { tree[rt].add=0; if(l==r) { tree[rt].sum=0; tree[rt].maxa=0; tree[rt].minb=b[l]; return; } int mid=(l+r)/2; build(l,mid,rt<<1); build(mid+1,r,rt<<1|1); push_up(rt); } void update(int L,int R,int l,int r,int rt) { if(l>=L&&r<=R) { tree[rt].maxa++; if(tree[rt].maxa<tree[rt].minb) { tree[rt].add++; return; } if(l==r&&tree[rt].maxa>=tree[rt].minb) { tree[rt].sum++; tree[rt].minb+=b[l]; return; } } push_down(rt); int mid=(l+r)/2; if(L<=mid) update(L,R,l,mid,rt<<1); if(R>mid) update(L,R,mid+1,r,rt<<1|1); push_up(rt); } int query(int L,int R,int l,int r,int rt) { if(l>=L&&r<=R) return tree[rt].sum; push_down(rt); int mid=(l+r)/2; int ans=0; if(L<=mid) ans+=query(L,R,l,mid,rt<<1); if(R>mid) ans+=query(L,R,mid+1,r,rt<<1|1); return ans; } int main() { while(~scanf("%d%d",&n,&Q)) { for(int i=1;i<=n;i++) scanf("%d",&b[i]); build(1,n,1); while(Q--) { char t[10];int x,y; scanf("%s%d%d",&t,&x,&y); if(t[0]=='a') update(x,y,1,n,1); else printf("%d\n",query(x,y,1,n,1)); } } return 0; }
