# hdu 6351

HDU 6315 Naive Operations(线段树区间整除区间)

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

# include

using namespace std;

const int maxn=(100000+10)*4;

# define Min(a,b) ((a)<(b)?(a):(b))

int b[maxn],add[maxn],minn[maxn],sum[maxn],n;

void pushup(int rt){
minn[rt]=Min(minn[rt<<1],minn[rt<<1|1]);
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}

void build(int l,int r,int rt){
sum[rt]=0,add[rt]=0;
if(l==r){
minn[rt]=b[l];
return ;
}
int mid=(l+r)>>1;
build(l,mid,rt<<1);
build(mid+1,r,rt<<1|1);
pushup(rt);
}

void pushdown(int rt){
if(add[rt]){
add[rt<<1|1]+=add[rt];
add[rt<<1]+=add[rt];
minn[rt<<1|1]-=add[rt];
minn[rt<<1]-=add[rt];
add[rt]=0;
}
}

void change(int l,int r,int rt,int L,int R){
if(l>=L && r<=R && minn[rt]>1){
add[rt]++;
minn[rt]--;
return ;
}
int mid=(l+r)>>1;
if(lr && minn[rt]1){
minn[rt]=b[l];
add[rt]=0;
sum[rt]++;
return ;
}
pushdown(rt);
if(L<=mid) change(l,mid,rt<<1,L,R);
if(R>mid) change(mid+1,r,rt<<1|1,L,R);
pushup(rt);
}

int ask(int l,int r,int rt,int L,int R){
if(l>=L && r<=R){
return sum[rt];
}
int mid=(l+r)>>1;
pushdown(rt);
int ans=0;
if(L<=mid) ans+=ask(l,mid,rt<<1,L,R);
if(R>mid) ans+=ask(mid+1,r,rt<<1|1,L,R);
pushup(rt);
return ans;
}

int main(){
int q,l,r;
while(scanf("%d%d",&n,&q)!=EOF){
for (int i=1;i<=n;i++) scanf("%d",&b[i]);
build(1,n,1);
while(q--){
char c[10];
scanf("%s",c);
if(c[0]=='a') {scanf("%d%d",&l,&r);change(1,n,1,l,r);}
else {scanf("%d%d",&l,&r);printf("%d\n",ask(1,n,1,l,r));}
}
}
return 0;
}

posted @ 2019-02-14 16:03  lmjer  阅读(179)  评论(0编辑  收藏  举报