Naive Operations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 502768/502768 K (Java/Others)
Total Submission(s): 2903 Accepted Submission(s): 1272
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
Source
2018 Multi-University Training Contest 2
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题解：
对于每个区间我们维护区间分子的最大值a，分母的最小值b，只有当区间最大的分子大于最小的分母时该区间分子都加1才有可能对答案有贡献，这个时候我们暴力的找到这个分子大于等于分母的点然后更新该区间的sum，并且让分母变成原来的两倍，这样做的意义在于只有当前节点再增加原来的分母次数的话才会对答案造成贡献。否则的话我们直接对当前区间打上标记后返回。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;
#define LL long long
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define sca(x) scanf("%d",&x);
#define N 100005
struct node
{
int a,b,sum,lazy;
}t[N<<2];
int c[N];
void pushup(int rt)
{
t[rt].sum=t[rt<<1].sum+t[rt<<1|1].sum;
t[rt].a=max(t[rt<<1].a,t[rt<<1|1].a);
t[rt].b=min(t[rt<<1].b,t[rt<<1|1].b);
}
void pushdown(int rt)
{
if(t[rt].lazy)
{
t[rt<<1].lazy+=t[rt].lazy;
t[rt<<1|1].lazy+=t[rt].lazy;
t[rt<<1].a+=t[rt].lazy;
t[rt<<1|1].a+=t[rt].lazy;
t[rt].lazy=0;
}
}
void build(int rt,int l,int r)
{
t[rt].sum=0;
t[rt].lazy=0;
if(l==r)
{
t[rt].b=c[l];
t[rt].a=0;
return ;
}
int m=(l+r)>>1;
build(rt<<1,l,m);
build(rt<<1|1,m+1,r);
pushup(rt);
}
void upda(int rt,int l,int r,int ql,int qr)
{
if(l>=ql && r<=qr)
{
t[rt].a++;
if(t[rt].a<t[rt].b)
{
t[rt].lazy++;
return ;
}
if(l==r && t[rt].a>=t[rt].b)
{
t[rt].sum++;
t[rt].b+=c[l];
return ;
}
}
pushdown(rt);
int m=(l+r)>>1;
if(ql<=m) upda(rt<<1,l,m,ql,qr);
if(qr>m) upda(rt<<1|1,m+1,r,ql,qr);
pushup(rt);
}
int query(int rt,int l,int r,int ql,int qr)
{
if(l>=ql && r<=qr)
{
return t[rt].sum;
}
pushdown(rt);
int m=(l+r)>>1;
int ans=0;
if(ql<=m) ans+=query(rt<<1,l,m,ql,qr);
if(qr>m) ans+=query(rt<<1|1,m+1,r,ql,qr);
return ans;
}
int main()
{
int n,m;
int l,r;
while(cin>>n>>m){
rep(i,1,n) sca(c[i]);
build(1,1,n);
char s[10];
while(m--)
{
scanf("%s%d%d",s,&l,&r);
if(s[0]=='a') upda(1,1,n,l,r);
else printf("%d\n",query(1,1,n,l,r));
}
}
}