ZOJ 3279-Ants(线段树)

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Ants

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Time Limit: 2 Seconds      Memory Limit: 32768 KB

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echo is a curious and clever girl, and she is addicted to the ants recently.

She knows that the ants are divided into many levels depends on ability, also, she finds the number of each level will change.

Now, she will give two kinds of operations as follow :

First, "p a b", the number of ants in level a change to b.

Second, "q x", it means if the ant's ability is rank xth in all ants, what level will it in?

Input

There are multi-cases, and you should use EOF to check whether it is in the end of the input. The first line is an integer n, means the number of level. (1 <= n <= 100000). The second line follows n integers, the ith integer means the number in level i. The third line is an integer k, means the total number of operations. Then following k lines, each line will be "p a b" or "q x", and 1 <= x <= total ants, 1 <= a <= n, 0 <= b. What's more, the total number of ants won't exceed 2000000000 in any time.

Output

Output each query in order, one query each line.

Sample Input

3
1 2 3
3
q 2
p 1 2
q 2

Sample Output

2
1

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Author: Lin, Yue
Source: ZOJ Monthly, December 2009

 

题意：

给你每个等级蚂蚁的数量 有两种操作：

（1）rank x的蚂蚁的等级是多少

（2）将等级为a的数量更改为b

题解：

还是一个简单的线段树，处理区间要仔细（调了一晚上bug-.-）

代码：

#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <utility>
#include <vector>
#define mem(arr, num) memset(arr, 0, sizeof(arr))
#define _for(i, a, b) for (int i = a; i <= b; i++)
#define __for(i, a, b) for (int i = a; i >= b; i--)
#define IO                     \
  ios::sync_with_stdio(false); \
  cin.tie(0);                  \
  cout.tie(0);
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f;
const double EPS = 1e-10;
const ll mod = 1000000007LL;
const int N = 1 << 19;
ll dat[N];
int num;
int tree_node;
void init(int n)
{
    tree_node = 1;
    while(tree_node < n)
        tree_node <<= 1;
    _for(i, 1, tree_node * 2 -1) dat[i] = 0;
}
void update(int pos, int date)
{
    pos += tree_node - 1;
    dat[pos] = date;
    while(pos > 1)
    {
        pos >>= 1;
        dat[pos] = dat[pos<<1] + dat[pos<<1|1];
    }
}
int query(int rank, int k)
{
    if(k > tree_node - 1)
        return k;
    if(rank <= dat[k<<1])
        return query(rank, k<<1);
    else
        return query(rank - dat[k<<1], k<<1|1);
}
void solve()
{
    ll x,p,q;
    char str;
    _for(i, 1, num)
    {
        scanf("%lld",&x);
        update(i,x);
    }
    scanf("%lld",&x);

    for(int i = 1; i <= x; i++)
    {
        getchar();
        scanf("%c",&str);
        if(str=='q')
        {
            scanf("%lld",&p);
            if(p > dat[1])
            {
                printf("%d\n",num);
            }
            else
                printf("%d\n",query(p,1) - tree_node + 1);
        }
        else
        {
            scanf("%lld%lld",&p,&q);
            update(p,q);
        }
    }
}
int main()
{

    while(scanf("%d",&num)!=EOF)
    {
        init(num);
        solve();
    }
    return 0;
}