SPOJ 2713(GSS4-线段树区间开方-多组数据)

创创精灵 lv.1

发布时间：2022-10-29 21:55:03 360

2713. Can you answer these queries IV Problem code: GSS4

Input

Multiple test cases, please proceed them one by one. Input terminates by EOF.

For each test case:

The first line contains an integer N. The following line contains N integers, representing the sequence A₁..A_N.
The third line contains an integer M. The next M lines contain the operations in the form "i x y".i=0 denotes the modify operation, i=1 denotes the query operation.

Output

For each test case:

Output the case number (counting from 1) in the first line of output. Then for each query, print an integer as the problem required.

Print an blank line after each test case.

See the sample output for more details.

Example

Input: 5 1 2 3 4 5 5 1 2 4 0 2 4 1 2 4 0 4 5 1 1 5 4 10 10 10 10 3 1 1 4 0 2 3 1 1 4 Output: Case #1: 9 4 6 Case #2: 40 26

这题我在交了《花神游历各国》以后直接Wa了……

定睛一看：

1.多组数据

2.a[i]取值≤10^18

3.l>r

4.数组初始化会T……

于是我T_Y

SPOJ 2713(GSS4-线段树区间开方-多组数据)_i++

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MAXN (300000+10)
#define MAXM (600000+10)
#define Lson (x<<1)
#define Rson ((x<<1)+1)
long long n,m,a[MAXN*2]={0},b[MAXN*2]={0};
long long sum[MAXN*2];
void update(int x)
{
  if (b[Lson]==2||b[Rson]==2) b[x]=2;
  else b[x]=1; 
  sum[x]=sum[Lson]+sum[Rson];
}
void build(int x,int l,int r)
{
  if (l==r) 
  {
    scanf("%lld",&a[x]);sum[x]=a[x];
    if (a[x]<=1) b[x]=1;
    else b[x]=2;
  }    
  if (l>=r) return;
  int m=(l+r)>>1;
  build(Lson,l,m);
  build(Rson,m+1,r);
  update(x);
}
void pushdown(int x,int l,int r)
{
  if (b[x]<=1) return;
  if (l==r)
  {
    a[x]=sum[x]=sqrt(a[x]);//floor(sqrt((double)a[x])+0.5);
    if (a[x]<=1) b[x]=1;
    return;
  }
  int m=(l+r)>>1;
  pushdown(Lson,l,m);
  if (m<r) pushdown(Rson,m+1,r);
  update(x);
}
void change(int x,int l,int r,int L,int R)
{
  if (b[x]<=1) return;
  int m=(l+r)>>1;
  if (L<=l&&r<=R)
  {
    pushdown(x,l,r);
    return;
  }
  if (L<=m) change(Lson,l,m,L,R);
  if (m<R) change(Rson,m+1,r,L,R);
  update(x);  
}
long long qur(int x,int l,int r,int L,int R)
{
  int m=(l+r)>>1;
  if (L<=l&&r<=R)
  {
    return sum[x];
  }
  long long ans=0;
  if (L<=m) ans+=qur(Lson,l,m,L,R);
  if (m<R) ans+=qur(Rson,m+1,r,L,R);
  return ans;  
}

int main()
{
//  freopen("spoj2713.in","r",stdin);
//  freopen(".out","w",stdout);
  int tt=1;
  while (scanf("%d",&n)!=-1)
  {
  //  memset(a,0,sizeof(a));
  //  memset(b,0,sizeof(b));
  //  memset(sum,0,sizeof(sum));
    printf("Case #%d:\n",tt++);
    build(1,1,n);
    scanf("%d",&m);
    For(i,m)
    {
      int x,l,r;
      scanf("%d%d%d",&x,&l,&r);
      if (l>r) swap(l,r);
      if (x==1) printf("%lld\n",qur(1,1,n,l,r));
      else change(1,1,n,l,r);
    }
    puts("");
  }
  return 0;
}

文章来源： https://blog.51cto.com/u_15724837/5793901

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SPOJ 2713(GSS4-线段树区间开方-多组数据)

2713. Can you answer these queries IV

Problem code: GSS4

Input

Output

Example