线段树和树状数组的全面配合与比较

线段树的考题，当然其实也可以用树状数组去解决的：

`One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.

The maze is organized as follows. Each room of the maze has two one-way portals. Let’s consider room number i(1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number pi, where 1 ≤ pi ≤ i.

In order not to get lost, Vasya decided to act as follows.

Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.

Let’s assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room pi), otherwise Vasya uses the first portal.

Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.

Input

The first line contains integer n (1 ≤ n ≤ 103) — the number of rooms. The second line contains n integers pi (1 ≤ pi ≤ i). Each pi denotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.

Output

Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007(109 + 7).

Sample Input

Input

2

1 2

Output

4

Input

4

1 1 2 3

Output

20

Input

5

1 1 1 1 1

Output

62“

这里我用下线段树去做下：

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<sstream>
#include<algorithm>
#include<utility>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#include<iterator>
#include<stack>
using namespace std;
#define e tree[id]
#define lson tree[id*2]
#define rson tree[id*2+1]
typedef __int64 LL;
const int INF=1e9+7;
const double eps=1e-7;
const int maxn=100005;
bool f[maxn];
int A[maxn];
int judge(int n)
{
int x=4;
while(n>0&&(x==4||x==7))
{
x=n%10;
n/=10;
}
if(x==4||x==7)return 1;
else return 0;
}
struct Tree
{
int le,ri;
int sum,d;
}tree[4*maxn];
void pushup(int id){ e.sum=lson.sum+rson.sum; }//取其和
void pushdown(int id)  //延迟更新
{
if(e.d!=0&&e.le!=e.ri) //d不为0且不是叶子节点
{
lson.sum+=(lson.ri-lson.le+1)*e.d; //左右儿子的和要加上他们自身的长度乘以d
rson.sum+=(rson.ri-rson.le+1)*e.d;
lson.d+=e.d; //推到下一层，因为子树还没更新
rson.d+=e.d;
e.d=0;   //更新完了得置为0，不然会重复更新
}
}
void Build_tree(int id,int le,int ri)  //建树
{
e.le=le,e.ri=ri,e.d=0;
if(le==ri){ e.sum=f[A[le]];return; }
int mid=(le+ri)/2;
Build_tree(id*2,le,mid);
Build_tree(id*2+1,mid+1,ri);
pushup(id);
}
void Update(int id,int x,int y,int d)  //更新
{
int le=e.le,ri=e.ri;
if(x<=le&&ri<=y){ e.sum+=f[e.sum+(ri-le+1)*d];e.d+=d; return; }
pushdown(id);  //推下去
int mid=(le+ri)/2;
if(x<=mid) Update(id*2,x,y,d);   //更新左边
if(y>mid)  Update(id*2+1,x,y,d); //更新右边
pushup(id);  //推上去
return;
}
//void update(int id,int x,int y)  //更新
//{
//    int le=e.le,ri=e.ri;
//    if(le==ri){ e.sum+=y;return;}
//    pushdown(id);  //推下去
//    int mid=(le+ri)/2;
//    if(x<=mid) update(id*2,x,y);   //更新左边
//    else  update(id*2+1,x,y); //更新右边
//    pushup(id);  //推上去
//    return;
//}
void update(int id,int k,int v)  //更新
{
int le=e.le,ri=e.ri;
if(le==ri){ e.sum+=v; return; }
pushdown(id);  //推下去
int mid=(le+ri)/2;
if(k<=mid) update(id*2,k,v);   //更新左边
else update(id*2+1,k,v); //更新右边
pushup(id);  //推上去
return;
}
int Query(int id,int x,int y)
{
int le=e.le,ri=e.ri;
if(x<=le&&ri<=y) return e.sum;
pushdown(id);   //这个一定要，不然查询结果会出错
int mid=(le+ri)/2;
int ret=0;
if(x<=mid) ret+=Query(id*2,x,y);
if(y>mid)  ret+=Query(id*2+1,x,y);
return ret;
}
bool jug(int a)
{
while(a)
{
if(a%10!=4&&a%10!=7) return 0;
a/=<
ec1f
span class="hljs-number">10;
}
return 1;
}
int main()
{

int N,Q,x,y,d;
char op[10];
memset(f,0,sizeof f);
for(int i=0;i<=10001;i++)
if(judge(i)) f[i]=1;
while(~scanf("%d%d",&N,&Q))
{
for(int i=1;i<=N;i++) scanf("%d",&A[i]);
Build_tree(1,1,N);
while(Q--)
{
scanf("%s",op);
if(op[0]=='a')
{
scanf("%d%d%d",&x,&y,&d);
for(int i=x;i<=y;i++)
{
int x1=f[A[i]];
A[i]+=d;
int x2=f[A[i]];
if(x2==1&&x1==0)update(1,i,1);
if(x1==1&&x2==0)update(1,i,-1)
;
}
}
else
{
scanf("%d%d",&x,&y);
printf("%d\n",Query(1,x,y));
}
}
}
return 0;
}
当然树状数组也可以写：

```#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <iostream>#define LL long long
using namespace std;

int a[100001],c[100002],kk[10002]={0};
int maxn;

inline int lowbit(int a)
{
return a&(-a);
}

void add(int x,int num)
{
while(x<=maxn)
{
c[x]+=num;
x+=lowbit(x);
}
}
int getsum(int x)
{
int sum=0;
while(x)
{
sum+=c[x];
x-=lowbit(x);
}
return sum;
}
bool jug(int a)
{
while(a)
{
if(a%10!=4&&a%10!=7) return 0;
a/=10;
}
return 1;
}
int main()
{
int i,ans,n,m,b,a1,c1,x1,x2;
char s[10];
for(i=1;i<=10001;i++)
if(jug(i)) kk[i]=1;
while(~scanf("%d %d",&n,&m))
{
maxn=n;
for(i=0; i<=n+1; i++)
c[i]=0;
for( i=1; i<=n; i++)
{
scanf("%d",&a[i]);
if(kk[a[i]]==1) add(i,1);
}
while(m--)
{
scanf("%s",s);
if(s[0]=='c')
{
scanf("%d %d",&a1,&b);
printf("%d\n",getsum(b)-getsum(a1-1));
}

else
{
scanf("%d %d %d",&a1,&b,&c1);
for(i=a1; i<=b; i++)
{
x1=kk[a[i]];
a[i]+=c1;
x2=kk[a[i]];
if(x1==1&&x2==0) add(i,-1);
else if(x1==0&&x2==1) add(i,1);
}
}
}
}
return 0;
}

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