[kuangbin带你飞]专题七线段树-白红宇

[kuangbin带你飞]专题七线段树

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发布时间：2019-06-12

本文共 22503 字，大约阅读时间需要 75 分钟。

C国的死对头A国这段时间正在进行军事演习，所以C国间谍头子Derek和他手下Tidy又开始忙乎了。A国在海岸线沿直线布置了N个工兵营地,Derek和Tidy的任务就是要监视这些工兵营地的活动情况。由于采取了某种先进的监测手段，所以每个工兵营地的人数C国都掌握的一清二楚,每个工兵营地的人数都有可能发生变动，可能增加或减少若干人手,但这些都逃不过C国的监视。

中央情报局要研究敌人究竟演习什么战术,所以Tidy要随时向Derek汇报某一段连续的工兵营地一共有多少人,例如Derek问:“Tidy,马上汇报第3个营地到第10个营地共有多少人!”Tidy就要马上开始计算这一段的总人数并汇报。但敌兵营地的人数经常变动，而Derek每次询问的段都不一样，所以Tidy不得不每次都一个一个营地的去数，很快就精疲力尽了，Derek对Tidy的计算速度越来越不满:"你个死肥仔，算得这么慢，我炒你鱿鱼!”Tidy想：“你自己来算算看，这可真是一项累人的工作!我恨不得你炒我鱿鱼呢!”无奈之下，Tidy只好打电话向计算机专家Windbreaker求救,Windbreaker说：“死肥仔，叫你平时做多点acm题和看多点算法书，现在尝到苦果了吧!”Tidy说："我知错了。。。"但Windbreaker已经挂掉电话了。Tidy很苦恼，这么算他真的会崩溃的，聪明的读者，你能写个程序帮他完成这项工作吗？不过如果你的程序效率不够高的话，Tidy还是会受到Derek的责骂的.

Input第一行一个整数T，表示有T组数据。

每组数据第一行一个正整数N（N<=50000）,表示敌人有N个工兵营地，接下来有N个正整数,第i个正整数ai代表第i个工兵营地里开始时有ai个人（1<=ai<=50）。

接下来每行有一条命令，命令有4种形式：

(1) Add i j,i和j为正整数,表示第i个营地增加j个人（j不超过30）

(2)Sub i j ,i和j为正整数,表示第i个营地减少j个人（j不超过30）;

(3)Query i j ,i和j为正整数,i<=j，表示询问第i到第j个营地的总人数;

(4)End 表示结束，这条命令在每组数据最后出现;

每组数据最多有40000条命令

Output对第i组数据,首先输出“Case i:”和回车,

对于每个Query询问，输出一个整数并回车,表示询问的段中的总人数,这个数保持在int以内。

Sample Input

1101 2 3 4 5 6 7 8 9 10Query 1 3Add 3 6Query 2 7Sub 10 2Add 6 3Query 3 10End

Sample Output

Case 1:63359 简单的线段树模板的应用，具体的细节可以看代码的注释。这题涉及到到区间求和和单点修改。

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long LL;const int maxn=50005;const int INF=0x3f3f3f3f;int n,p,q,c;int a[maxn];char s[10];struct Tree//存储这个线段的端点值和线段维护的信息{    int l,r;    int sum;}tree[maxn<<2];void build(int k,int l,int r)//建树的过程{    tree[k].l=l;tree[k].r=r;    if(l==r)    {        tree[k].sum=a[l];        return;    }    int mid=(l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);    tree[k].sum=tree[k<<1].sum+tree[k<<1|1].sum;}void change(int k,int l,int r,int val)/*更新的过程，我习惯性将区间更新和单点更新记一起，区间更新就是多了一个pushdown，单点更新的时候只需要在传参数的时候把l，r传同一个值就可以了。*/{    if(tree[k].l==l&&tree[k].r==r)    {        tree[k].sum+=val;        return;    }    if(tree[k].l==tree[k].r)        return;    int mid=(tree[k].l+tree[k].r)>>1;    if(mid>=r)        change(k<<1,l,r,val);    else if(mid
          
           >1; if(r<=mid) ans+=query(k<<1,l,r); else if(l>=mid+1) ans+=query(k<<1|1,l,r); else { ans+=query(k<<1,l,mid); ans+=query(k<<1|1,mid+1,r); } return ans;}int main(){ int T,casee=1; scanf("%d",&T); while(T--) { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); memset(tree,0,sizeof(tree)); build(1,1,n); printf("Case %d:\n",casee++); while(scanf("%s",s)) { if(s[0]=='E') break; else if(s[0]=='A') { scanf("%d %d",&p,&c); change(1,p,p,c); } else if(s[0]=='S') { scanf("%d %d",&p,&c); change(1,p,p,-c); } else if(s[0]=='Q') { scanf("%d %d",&p,&q); printf("%d\n",query(1,p,q)); } } } return 0;}

很多学校流行一种比较的习惯。老师们很喜欢询问，从某某到某某当中，分数最高的是多少。

这让很多学生很反感。

不管你喜不喜欢，现在需要你做的是，就是按照老师的要求，写一个程序，模拟老师的询问。当然，老师有时候需要更新某位同学的成绩。

Input本题目包含多组测试，请处理到文件结束。  在每个测试的第一行，有两个正整数 N 和 M ( 0
    
     <=200000,0
     
      <5000 )，分别代表学生的数目和操作的数目。  学生ID编号分别从1编到N。  第二行包含N个整数，代表这N个学生的初始成绩，其中第i个数代表ID为i的学生的成绩。  接下来有M行。每一行有一个字符 C (只取'Q'或'U') ，和两个正整数A，B。  当C为'Q'的时候，表示这是一条询问操作，它询问ID从A到B(包括A,B)的学生当中，成绩最高的是多少。  当C为'U'的时候，表示这是一条更新操作，要求把ID为A的学生的成绩更改为B。  Output对于每一次询问操作，在一行里面输出最高成绩。Sample Input

5 61 2 3 4 5Q 1 5U 3 6Q 3 4Q 4 5U 2 9Q 1 5

Sample Output

5659 区间查询和单点更新

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long LL;const int maxn=200005;const int INF=0x3f3f3f3f;int n,m,p,q,c;int a[maxn];char s[10];struct Tree{    int l,r;    int sum;}tree[maxn<<2];void build(int k,int l,int r){    tree[k].l=l;tree[k].r=r;    if(l==r)    {        tree[k].sum=a[l];        return;    }    int mid=(l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);    tree[k].sum=max(tree[k<<1].sum,tree[k<<1|1].sum);}void change(int k,int l,int r,int val){    if(tree[k].l==l&&tree[k].r==r)    {        tree[k].sum=val;        return;    }    if(tree[k].l==tree[k].r)        return;    int mid=(tree[k].l+tree[k].r)>>1;    if(mid>=r)        change(k<<1,l,r,val);    else if(mid
          
           >1; if(r<=mid) ans=query(k<<1,l,r); else if(l>=mid+1) ans=query(k<<1|1,l,r); else { ans=max(query(k<<1,l,mid),query(k<<1|1,mid+1,r)); } return ans;}int main(){ while(scanf("%d %d",&n,&m)!=EOF) { for(int i=1;i<=n;i++) scanf("%d",&a[i]); memset(tree,0,sizeof(tree)); build(1,1,n); while(m--) { scanf("%s",s); if(s[0]=='Q') { scanf("%d %d",&p,&q); printf("%d\n",query(1,p,q)); } else { scanf("%d %d",&p,&c); change(1,p,p,c); } } } return 0;}

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.

The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.

Each of the next Q lines represents an operation.

"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.

"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915 区间更新和区间查询，要用到lazy标记。 题意：支持两种操作，一个是把一个区间内每个人都加上或都减去同一个数，一个是查询一个区间内的最大值。

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long LL;const int maxn=100005;const int INF=0x3f3f3f3f;int n,m,p,q,c;int a[maxn];char s[10];struct Tree{    int l,r;    LL sum,lazy;}tree[maxn*8];void build(int k,int l,int r){    tree[k].l=l;tree[k].r=r;tree[k].lazy=0;    if(l==r)    {        tree[k].sum=a[l];        return;    }    int mid=(l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);    tree[k].sum=tree[k<<1].sum+tree[k<<1|1].sum;}void pushdown(int x){    if(tree[x].lazy!=0)    {        tree[x<<1].lazy+=tree[x].lazy;        tree[x<<1|1].lazy+=tree[x].lazy;        tree[x<<1].sum+=tree[x].lazy*(tree[x<<1].r-tree[x<<1].l+1);        tree[x<<1|1].sum+=tree[x].lazy*(tree[x<<1|1].r-tree[x<<1|1].l+1);        tree[x].lazy=0;    }}void change(int k,int l,int r,int val){    pushdown(k);    if(tree[k].l==l&&tree[k].r==r)    {        tree[k].lazy+=val;        tree[k].sum+=val*(r-l+1);        return;    }    if(tree[k].l==tree[k].r)        return;    int mid=(tree[k].l+tree[k].r)>>1;    if(mid>=r)        change(k<<1,l,r,val);    else if(mid
          
           >1; if(r<=mid) ans+=query(k<<1,l,r); else if(l>=mid+1) ans+=query(k<<1|1,l,r); else { ans+=query(k<<1,l,mid); ans+=query(k<<1|1,mid+1,r); } return ans;}int main(){ while(scanf("%d %d",&n,&m)!=EOF) { for(int i=1;i<=n;i++) scanf("%d",&a[i]); memset(tree,0,sizeof(tree)); build(1,1,n); while(m--) { scanf("%s",s); if(s[0]=='Q') { scanf("%d %d",&p,&q); printf("%lld\n",query(1,p,q)); } else { scanf("%d %d %d",&p,&q,&c); change(1,p,q,c); } } } return 0;}

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules:

Every candidate can place exactly one poster on the wall.

All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown).

The wall is divided into segments and the width of each segment is one byte.

Each poster must completely cover a contiguous number of wall segments.

They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections.

Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall.

Input

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among the n lines contains two integer numbers l _i and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= l _i <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered l _i, l _i+1 ,... , ri.

Output

For each input data set print the number of visible posters after all the posters are placed.

The picture below illustrates the case of the sample input.

Sample Input

151 42 68 103 47 10

Sample Output

4 区间覆盖，但是这题需要离散化。

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long LL;const int maxn=10005;const int INF=0x3f3f3f3f;int n,m,p,q,c;int li[maxn],ri[maxn];int x[maxn<<1],hashh[10000010];struct Tree{    int l,r;    int lazy;}tree[maxn<<3];void build(int k,int l,int r){    tree[k].l=l;tree[k].r=r;tree[k].lazy=0;    if(l==r)    {//        tree[k].sum=a[l];        return;    }    int mid=(l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);//    tree[k].sum=tree[k<<1].sum+tree[k<<1|1].sum;}//void pushdown(int x)//{//    if(tree[x].l!=tree[x].r)//    {//        tree[x<<1].lazy+=tree[x].lazy;//        tree[x<<1|1].lazy+=tree[x].lazy;//        tree[x<<1].sum+=tree[x].lazy*(tree[x<<1].r-tree[x<<1].l+1);//        tree[x<<1|1].sum+=tree[x].lazy*(tree[x<<1|1].r-tree[x<<1|1].l+1);//    }//    tree[x].lazy=0;//}bool change(int k,int l,int r){    if(tree[k].lazy)        return false;    if(tree[k].l==l&&tree[k].r==r)    {        tree[k].lazy=true;//        tree[k].sum+=val*(r-l+1);        return true;    }    bool temp;    int mid=(tree[k].l+tree[k].r)>>1;    if(mid>=r)        temp=change(k<<1,l,r);    else if(mid
          
           >1;// if(r<=mid)// ans+=query(k<<1,l,r);// else if(l>=mid+1)// ans+=query(k<<1|1,l,r);// else// {// ans+=query(k<<1,l,mid);// ans+=query(k<<1|1,mid+1,r);// }// return ans;//}int main(){ int T; scanf("%d",&T); while(T--) { scanf("%d",&n); memset(tree,0,sizeof(tree)); int cnt=0; for(int i=0;i
           
            =0;i--) { if(change(1,hashh[li[i]],hashh[ri[i]])) ans++; } printf("%d\n",ans); } return 0;}

In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.

Now Pudge wants to do some operations on the hook.

Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.

The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:

For each cupreous stick, the value is 1.

For each silver stick, the value is 2.

For each golden stick, the value is 3.

Pudge wants to know the total value of the hook after performing the operations.

You may consider the original hook is made up of cupreous sticks.

InputThe input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.  For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.  Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.  OutputFor each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.  Sample Input

11021 5 25 9 3

Sample Output

Case 1: The total value of the hook is 24.

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long LL;const int maxn=10005;const int INF=0x3f3f3f3f;int n,m,p,q,c;int li[maxn],ri[maxn];int x[maxn<<1],hashh[10000010];struct Tree{    int l,r;    LL sum,lazy;}tree[maxn<<3];void build(int k,int l,int r){    tree[k].l=l;tree[k].r=r;tree[k].lazy=0;    if(l==r)    {        tree[k].sum=1;        return;    }    int mid=(l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);    tree[k].sum=tree[k<<1].sum+tree[k<<1|1].sum;}void pushdown(int x){    if(tree[x].l!=tree[x].r)    {        tree[x<<1].lazy=tree[x].lazy;        tree[x<<1|1].lazy=tree[x].lazy;        tree[x<<1].sum=tree[x].lazy*(tree[x<<1].r-tree[x<<1].l+1);        tree[x<<1|1].sum=tree[x].lazy*(tree[x<<1|1].r-tree[x<<1|1].l+1);    }    tree[x].lazy=0;}void change(int k,int l,int r,int val){    if(tree[k].lazy)        pushdown(k);    if(tree[k].l==l&&tree[k].r==r)    {        tree[k].sum=val*(r-l+1);        tree[k].lazy=val;        return;    }    int mid=(tree[k].l+tree[k].r)>>1;    if(mid>=r)        change(k<<1,l,r,val);    else if(mid
          
           >1; if(r<=mid) ans+=query(k<<1,l,r); else if(l>=mid+1) ans+=query(k<<1|1,l,r); else { ans+=query(k<<1,l,mid); ans+=query(k<<1|1,mid+1,r); } return ans;}int main(){ int T,casee=1; scanf("%d",&T); while(T--) { scanf("%d",&n); memset(tree,0,sizeof(tree)); build(1,1,n); scanf("%d",&m); while(m--) { scanf("%d %d %d",&p,&q,&c); change(1,p,q,c); } printf("Case %d: The total value of the hook is %d.\n",casee++,query(1,1,n)); } return 0;}

Painting some colored segments on a line, some previously painted segments may be covered by some the subsequent ones.

Your task is counting the segments of different colors you can see at last.

Input The first line of each data set contains exactly one integer n, 1 <= n <= 8000, equal to the number of colored segments.

Each of the following n lines consists of exactly 3 nonnegative integers separated by single spaces:

x1 x2 c

x1 and x2 indicate the left endpoint and right endpoint of the segment, c indicates the color of the segment.

All the numbers are in the range [0, 8000], and they are all integers.

Input may contain several data set, process to the end of file.

Output Each line of the output should contain a color index that can be seen from the top, following the count of the segments of this color, they should be printed according to the color index.

If some color can't be seen, you shouldn't print it.

Print a blank line after every dataset.

Sample Input 5 0 4 4 0 3 1 3 4 2 0 2 2 0 2 3 4 0 1 1 3 4 1 1 3 2 1 3 1 6 0 1 0 1 2 1 2 3 1 1 2 0 2 3 0 1 2 1

Sample Output 1 1 2 1 3 1

1 1

0 2

1 1

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long LL;const int maxn=8005;const int INF=0x3f3f3f3f;int n,m,p,q,c;int a[maxn],ans[maxn];struct Tree{    int l,r;    int lazy;}tree[maxn<<2];void build(int k,int l,int r){    tree[k].l=l;tree[k].r=r;tree[k].lazy=-1;    if(l==r)    {        return;    }    int mid=(l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);}void pushdown(int x){    if(tree[x].l!=tree[x].r)    {        tree[x<<1].lazy=tree[x].lazy;        tree[x<<1|1].lazy=tree[x].lazy;        tree[x].lazy=-1;    }}void change(int k,int l,int r,int val){    if(tree[k].lazy!=-1)        pushdown(k);    if(tree[k].l==l&&tree[k].r==r)    {        tree[k].lazy=val;        return;    }    int mid=(tree[k].l+tree[k].r)>>1;    if(mid>=r)        change(k<<1,l,r,val);    else if(mid
          
           >1; if(r<=mid) query(k<<1,l,r); else if(l>=mid+1) query(k<<1|1,l,r); else { query(k<<1,l,mid); query(k<<1|1,mid+1,r); }}int main(){ while(scanf("%d",&n)!=EOF) { memset(tree,0,sizeof(tree)); memset(ans,0,sizeof(ans)); memset(a,-1,sizeof(a)); build(1,1,8001); for(int i=0;i

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, N and Q.

Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i

Lines N+2.. N+ Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Output

Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

6 31734251 54 62 2

Sample Output

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long LL;const int maxn=50005;const int INF=0x3f3f3f3f;int n,m,p,q,c;int a[maxn];int ansmax,ansmin;struct Tree{    int l,r;    int maxx,minn;}tree[maxn<<3];void build(int k,int l,int r){    tree[k].l=l;tree[k].r=r;tree[k].minn=INF;tree[k].maxx=0;    if(l==r)    {        tree[k].maxx=a[l];        tree[k].minn=a[l];        return;    }    int mid=(l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);    tree[k].minn=min(tree[k<<1].minn,tree[k<<1|1].minn);    tree[k].maxx=max(tree[k<<1].maxx,tree[k<<1|1].maxx);}void query(int k,int l,int r){    if(tree[k].l==l&&tree[k].r==r)    {        ansmax=max(ansmax,tree[k].maxx);        ansmin=min(ansmin,tree[k].minn);        return;    }    int mid=(tree[k].l+tree[k].r)>>1;    if(r<=mid)        query(k<<1,l,r);    else if(l>=mid+1)        query(k<<1|1,l,r);    else    {        query(k<<1,l,mid);        query(k<<1|1,mid+1,r);    }}int main(){    while(scanf("%d %d",&n,&m)!=EOF)    {        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        memset(tree,0,sizeof(tree));        build(1,1,n);        while(m--)        {            scanf("%d %d",&p,&q);            ansmax=0;ansmin=INF;            query(1,p,q);            printf("%d\n",ansmax-ansmin);        }    }    return 0;}

A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help.

You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line.

Notice that the square root operation should be rounded down to integer.

InputThe input contains several test cases, terminated by EOF.    For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000)    The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 2 63.    The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000)    For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive.  OutputFor each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.Sample Input

101 2 3 4 5 6 7 8 9 1050 1 101 1 101 1 50 5 81 4 8

Sample Output

Case #1:1976

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long LL;const int maxn=100005;const int INF=0x3f3f3f3f;int n,m,p,q,flag;LL a[maxn],ans[maxn];struct Tree{    int l,r;    LL sum,len;}tree[maxn<<2];void build(int k,int l,int r){    tree[k].l=l;tree[k].r=r;tree[k].len=r-l+1;    if(l==r)    {        tree[k].sum=a[l];        return;    }    int mid=(l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);    tree[k].sum=tree[k<<1].sum+tree[k<<1|1].sum;}void pushdown(int k,int l,int r){    if(tree[k].l==tree[k].r)    {        tree[k].sum=(LL)sqrt(tree[k].sum);        return;    }    int mid=(tree[k].l+tree[k].r)>>1;    pushdown(k<<1,l,mid);    pushdown(k<<1|1,mid+1,r);    tree[k].sum=tree[k<<1].sum+tree[k<<1|1].sum;}void change(int k,int l,int r){    if(tree[k].l==l&&tree[k].r==r)    {        if(tree[k].sum==tree[k].len)            return;        pushdown(k,l,r);        return;    }    int mid=(tree[k].l+tree[k].r)>>1;    if(mid>=r)        change(k<<1,l,r);    else if(mid
          
           >1; if(r<=mid) ans+=query(k<<1,l,r); else if(l>=mid+1) ans+=query(k<<1|1,l,r); else { ans+=query(k<<1,l,mid); ans+=query(k<<1|1,mid+1,r); } return ans;}int main(){ int casee=1; while(scanf("%d",&n)!=EOF) { for(int i=1;i<=n;i++) scanf("%lld",&a[i]); build(1,1,n); scanf("%d",&m); printf("Case #%d:\n",casee++); while(m--) { scanf("%d %d %d",&flag,&p,&q); if(p>q) swap(p,q); if(!flag) change(1,p,q); else printf("%lld\n",query(1,p,q)); } printf("\n"); } return 0;}