zoukankan      html  css  js  c++  java
  • ZOJ 3279 Ants

     

    Ants

    Time Limit: 2 Seconds      Memory Limit: 32768 KB

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



    /*
      ZOJ 3279
    */

    #include <iostream>
    #include <cstdio>
    #include <cstring>

    using namespace std;

    const int maxn=100000;

    int Tree[100010];
    int k[100010];
    int lowbit(int x)
    {
        return -x&x;
    }

    void add(int x,int val)
    {
        for(int i=x;i<=maxn;i+=lowbit(i))
            Tree+=val;
    }

    int get(int x)
    {
        int sum=0;
        for(int i=x;i;i-=lowbit(i))
            sum+=Tree;
        return sum;
    }
    int n,m;
    int main()
    {
    while(scanf("%d",&n)!=EOF)
    {
        memset(Tree,0,sizeof(Tree));

        for(int i=1;i<=n;i++)
        {
            scanf("%d",&k);
            add(i,k);
        }
        scanf("%d",&m);
        char ch[2];int a,b;
        while(m--)
        {
            scanf("%s",ch);
            if(ch[0]=='p')
            {
                scanf("%d%d",&a,&b);
                add(a,-k[a]);
                add(a,b);
                k[a]=b;
            }
            if(ch[0]=='q')
            {
                scanf("%d",&a);
                int high=n,low=1;
                int ans=n;
                while(low<=high)
                {
                   int mid=(high+low)/2;
                   if(get(mid)>=a)
                   {
                       high=mid-1;
                       ans=mid;
                   }
                   else
                   {
                       low=mid+1;
                   }
                }
                printf("%d ",ans);
            }
        }
    }

        return 0;
    }
    * This source code was highlighted by YcdoiT. ( style: Codeblocks )
  • 相关阅读:
    ViScript 1.0 Released
    How to: 修改程序的拖拽行为
    API Hooking 的原理
    小T历险记
    我的酒窝.NET
    MSN Space
    Naive Container 发布1.0版本
    EFT acceptance and functional testing tool for Windows application
    [译]JavaScript:如何判断值的类型
    [译]JavaScript:多行字符串
  • 原文地址:https://www.cnblogs.com/CKboss/p/3350896.html
Copyright © 2011-2022 走看看