Naive Operations
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 502768/502768 K (Java/Others)
Total Submission(s): 1205 Accepted Submission(s): 500
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
题意：给一个b数组，为n的全排列，有一个初始化为零的a数组，有两种操作：
add x y 表示a数组中区间为[x,y]的元素全部加1
query x y 表示查找区间为[x,y]的a[i]/b[i]之和
可知每当a[i]加了b[i]次时a[i]/b[i]的值加一，可以维护一个线段树，表示当前区间内加1的情况，每当a[i]增加了b[i]次时更新，可以求得最多更新10000+10000/2+10000/3+...+1次。在维护一个树状数组求出答案。
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<queue>
#include<vector>
using namespace std;
#define ll long long
const int maxm = 100005;
struct node
{
int l, r, mi, val;
}tr[maxm*8];
int c[maxm], b[maxm], n;
char str[maxm];
void add(int k, int val)
{
while (k <= n)
c[k] += val, k += k&-k;
}
int query(int k)
{
int ans = 0;
while (k)
ans += c[k], k -= k&-k;
return ans;
}
void make(int l, int r, int num)
{
tr[num].l = l, tr[num].r = r;
tr[num].val = 0;
if (l == r)
{
tr[num].mi = b[l], tr[num].val = 0;
return;
}
int mid = (tr[num].l + tr[num].r) / 2;
make(l, mid, num * 2);
make(mid + 1, r, num * 2 + 1);
tr[num].mi = min(tr[num * 2].mi, tr[num * 2 + 1].mi);
}
void updown(int num)
{
tr[num * 2].val += tr[num].val;
tr[num * 2 + 1].val += tr[num].val;
tr[num * 2].mi -= tr[num].val;
tr[num * 2 + 1].mi -= tr[num].val;
tr[num].val = 0;
}
void update(int l, int r, int num,int val)
{
if (tr[num].l == l&&tr[num].r == r || val == 0)
{
tr[num].val += val, tr[num].mi -= val;
if (tr[num].mi > 0) return;
if (l != r)
{
updown(num);
int mid = (tr[num].l + tr[num].r) / 2;
update(l, mid, num * 2, 0);
update(mid + 1, r, num * 2 + 1, 0);
tr[num].mi = min(tr[num * 2].mi, tr[num * 2 + 1].mi);
}
else
{
tr[num].mi = b[l], tr[num].val = 0;
add(l, 1);
}
return;
}
updown(num);
int mid = (tr[num].l + tr[num].r) / 2;
if (l > mid) update(l, r, num * 2 + 1, 1);
else if (r <= mid) update(l, r, num * 2, 1);
else
{
update(l, mid, num * 2, 1);
update(mid + 1, r, num * 2 + 1, 1);
}
tr[num].mi = min(tr[num * 2].mi, tr[num * 2 + 1].mi);
}
int main()
{
int i, j, k, sum, m, x, y;
while (scanf("%d%d", &n, &m) != EOF)
{
for (i = 1;i <= n;i++)
{
scanf("%d", &b[i]);
c[i] = 0;
}
make(1, n, 1);
for (i = 1;i <= m;i++)
{
scanf("%s%d%d", str, &x, &y);
if (str[0] == 'a')
update(x, y, 1, 1);
else printf("%d\n", query(y) - query(x - 1));
}
}
return 0;
}