nefu 三国之战

第十四届蓝桥杯大赛软件组省赛 Python大学A组 个人暴力题解 A, B, C, D, E, F, G, H, I, J 阅读详情

三国之战
Time Limit 1000 Memory Limit 65536
 
description
  在三国某某战争中,魏军要围攻蜀国. ,诸葛亮亲自指挥战争,对抗曹操。现假设诸葛亮有N个大炮,每个大炮都有一个攻击值ai(ai>=0 &&ai<=100) ,  而且他有一个神奇的技能,就是他能指定任意区间的大炮,使他们的攻击值都乘以x(x.>=0 && x<=10),同时他也会询问某段区间的大炮的总攻击值。(数据保证答案不会超过long long)

 
input
   第一行输入N(N<=100000),接下一行N个数ai,代表第i个大炮的攻击值
   接着输入Q,代表Q次操作,格式:
   attack  l  r  x  代表把第l到 第r个大炮的攻击值都乘以x
   query  l  r   查询l到r大炮攻击和值
 
output
   输出每次查询结果
 
sample_input
4
1 1 1 1
3
attack 1 2 2
query  1 4
query  2 2
 

3
100 100 100
3
query 1 3
attack  1 3 0
query 1 3
 
sample_output
Case 1:
6
2
Case 2:
300
0
 
hint

 
source

这一题我A了很久才搞定,主要时间就花在调试上面
因为一个漏洞。

6
1 1 1 1 1 1
6
attack 1 4 4
attack 3 6 2
attack 1 1 3
attack 1 5 3
attack 1 4 0
query 3 6

8

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 100030
using namespace std;
class node
{
  public:
  int l,r;
  int sum,lazy;
  bool flag;
}root[4*N];
int shu[N];
void build(int t,int l,int r)
{
  root[t].l=l;root[t].r=r;root[t].lazy=1;
  root[t].flag=0;
  if(root[t].l==root[t].r){cin>>root[t].sum;return;}
  build(t<<1,l,(l+r)/2);build(t<<1|1,(l+r)/2+1,r);
  root[t].sum=root[t<<1].sum+root[t<<1|1].sum;
}
void up(int t)
{
  if(root[t<<1].flag&&root[t<<1].l!=root[t<<1].r)up(t<<1);
  if(root[t<<1|1].flag&&root[t<<1|1].l!=root[t<<1|1].r)up(t<<1|1);
  root[t<<1].lazy*=root[t].lazy;root[t<<1|1].lazy*=root[t].lazy;
      root[t<<1].flag=1;root[t<<1|1].flag=1;
      root[t<<1].sum*=root[t].lazy;
      root[t<<1|1].sum*=root[t].lazy;
      root[t].lazy=1;
      root[t].flag=0;
}
void update(int t,int l,int r,int date)
{
  if(root[t].l==l&&root[t].r==r)
  { if(root[t].flag&&root[t].l!=root[t].r)//就是这调了N久才知道要补上
    up(t);
    root[t].sum*=date;root[t].lazy=date;root[t].flag=1; return;}
  else
  {
    if(root[t].flag&&root[t].l!=root[t].r)
    {
      up(t);
    }
    if(root[t<<1].r>=r)update(t<<1,l,r,date);
    else if(root[t<<1|1].l<=l)update(t<<1|1,l,r,date);
    else {update(t<<1,l,root[t<<1].r,date);update(t<<1|1,root[t<<1|1].l,r,date);}
    root[t].sum=root[t<<1].sum+root[t<<1|1].sum;
  }
}
int query(int t,int l,int r)
{
  int s;
  if(root[t].l==l&&root[t].r==r)return root[t].sum;
  else
  {
    if(root[t].flag&&root[t].l!=root[t].r)
    {
      up(t);
    }
   if(root[t<<1].r>=r)s=query(t<<1,l,r);
   else if(root[t<<1|1].l<=l)
   s=query(t<<1|1,l,r);
   else s=query(t<<1,l,root[t<<1].r)+query(t<<1|1,root[t<<1|1].l,r);
    root[t].sum=root[t<<1].sum+root[t<<1|1].sum;
  }
  return s;
}
int main()
{
  int n,q;
  char s[33];
  int a,b,c;
  int cas=0;
  while(scanf("%d",&n)!=EOF)
  {
    build(1,1,n);
    scanf("%d",&q);
    printf("Case %d:\n",++cas);
    //cout<<"Case "<<++cas<<": "<<"\n";
    for(int j=1;j<=q;j++)
    {  //cin>>s>>a>>b;
       scanf("%s%d%d",s,&a,&b);
      if(strcmp(s,"attack")==0)
     {
      scanf("%d",&c);
      update(1,a,b,c);
     }
    else
     {
       //cout<<query(1,a,b)<<"\n";
       printf("%d\n",query(1,a,b));
     }
    }

  }

}

深入理解DQN中的Q值:核心概念和应用 在强化学习中,Q值(或称为动作价值函数)是核心概念之一。Q值函数 ( Q(s, a) ) 表示在给定状态 ( s ) 下,采取动作 ( a ) 并遵循某策略后,智能体从当前状态到未来状态所能获得的预期总回报。简言之,Q值是评估特定状态和动作对的好坏的指标。 阅读详情

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nn.output = 'softmax'; % use softmax outputopts.numepochs = 5; % Number of full sweeps through dataopts.batchsize = 1000; % Take a mean gradient step over this many samplesopts.plot = 1; % enable plottingnn = nntrain(nn, tx, ty, opts, vx, vy); % nntrain takes validation set as last two arguments (optionally)[er, bad] = nntest(nn, test_x, test_y);assert(er < 0.1, 'Too big error');```[![Bitdeli Badge](https://d2weczhvl823v0.cloudfront.net/rasmusbergpalm/deeplearntoolbox/trend.png)](https://bitdeli.com/free "Bitdeli Badge")

codeforce 356 A. Knight Tournament(线段树覆盖,3级)

A. Knight Tournament time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Hooray! Berl II, the king of Berland i

nealgavin的专栏 1284

nefu700 Car race game(离散化+线段树)

Car race game Time Limit 1000ms Memory Limit 65536K description   Bob is a game programming specialist. In his new car race game, there are some r

nealgavin的专栏 1147

NEFU 708 Longest Non-decreasing Substring(线段树+DP)

Longest Non-decreasing Substring Time Limit 1000ms Memory Limit 65536K description You are given a string S no longer than 100,000 consisting of c

nealgavin的专栏 820

hdu 1556 Color the ball(线段树+懒惰标记)

Color the ball Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4961    Accepted Submission(s): 2624 Problem Description N个气球排成一排,从左

nealgavin的专栏 792

线段树的分析理解(不定期更新)

基础知识点; 所谓线段树,也就是一颗树,不过这棵树数管的是一个线段,而原本的树管的只是不同的点,其实本质上是一个道理。 线段树算法主要由三部分组成,1建树,2更新3查询。 就以其中第一题为例子 #define lson t #define rson t #define midl (l+r)/2 #define midr (l+r)/2+1 using namespace std;

nealgavin的专栏 719

hdu 4638 Group(线段树,离线维护左边界,4级)

Group Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 369    Accepted Submission(s): 207 Problem Description There are n men ,every man

nealgavin的专栏 688

hdu 4339 Query(线段树+二分,4级)

Query Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1956    Accepted Submission(s): 670 Problem Description You are given two s

nealgavin的专栏 685

hdu 4302 Holedox Eating(线段树或MAP,SET,4级)

Holedox Eating Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2341    Accepted Submission(s): 792 Problem Description Holedox is a sma

nealgavin的专栏 670

poj 2299 Ultra-QuickSort(求逆序数,树状数组)

Ultra-QuickSort Time Limit: 7000MS   Memory Limit: 65536K Total Submissions: 27736   Accepted: 9946 Description In this problem, you have to analyze a particular sorting al

nealgavin的专栏 625
上一篇: poj 3468 A Simple Problem with Integers(线段树)
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