codeforces 121 E. Lucky Array

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya has an array consisting of n numbers. He wants to perform m operations of two types:

• add l r d — add an integer d to all elements whose indexes belong to the interval from l to r, inclusive (1 ≤ l ≤ r ≤ n, 1 ≤ d ≤ 104);
• count l r — find and print on the screen how many lucky numbers there are among elements with indexes that belong to the interval from l to r inclusive (1 ≤ l ≤ r ≤ n). Each lucky number should be counted as many times as it appears in the interval.

Petya has a list of all operations. The operations are such that after all additions the array won't have numbers that would exceed 104. Help Petya write a program that would perform these operations.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 105) — the number of numbers in the array and the number of operations correspondingly. The second line contains n positive integers, none of which exceeds 104 — those are the array numbers. Next m lines contain operations, one per line. They correspond to the description given in the statement.

It is guaranteed that after all operations are fulfilled each number in the array will not exceed 104.

Output

For each operation of the second type print the single number on the single line — the number of lucky numbers in the corresponding interval.

Examples

input

3 6
2 3 4
count 1 3
count 1 2
add 1 3 2
count 1 3
add 2 3 3
count 1 3

output

1
0
1
1

input

4 5
4 4 4 4
count 1 4
add 1 4 3
count 1 4
add 2 3 40
count 1 4

output

4
4
4

Note

In the first sample after the first addition the array will look in the following manner:

4 5 6

After the second addition:

4 8 9

The second sample after the first addition:

7 7 7 7

After the second addition:

7 47 47 7

线段树区间修改，单点查询 

if(v) single_change(root,i,v);

这一句没写 T成傻逼了。

指针线段树比数组线段树慢近300ms

数组线段树比树状数组慢近1000ms

屠龙宝刀点击就送

 线段树代码

#include <ctype.h>
#include <cstdio>

const int N = 1e6+5;
bool IsLucky[N];
int dis[N],n,m,Lucky[50]={0,4,7,44,47,74,77,444,447,474,477,744,747,774,777,4444,4447,4474,4477,4744,4747,4774,4777,7444,7447,7474,7477,7744,7747,7774,7777};
inline void Read(int &x)
{
    bool f=0;
    register char ch=getchar();
    for(x=0;!isdigit(ch);ch=getchar()) if(ch=='-') f=1;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
    x=f?-x:x;
}
struct Segment
{
    int l,r,mid,flag,upval;
    Segment * ch[2];
    Segment()
    {
        ch[0]=ch[1]=NULL;
        flag=upval=0;
    }
};
inline void pushup(Segment *&k) {k->upval=k->ch[0]->upval+k->ch[1]->upval;}
void build(Segment *&k,int l,int r)
{
    k=new Segment;
    k->l=l;k->r=r;
    if(l==r)
    {
        if(IsLucky[dis[l]]) k->upval+=1;
        return;
    }
    k->mid=l+r>>1;
    build(k->ch[0],l,k->mid);
    build(k->ch[1],k->mid+1,r);
    pushup(k);
}
int query(Segment *&k,int l,int r)
{
    if(k->l==l&&k->r==r)
        return k->upval;
    if(l>k->mid) return query(k->ch[1],l,r);
    else if(r<=k->mid) return query(k->ch[0],l,r);
    else return query(k->ch[0],l,k->mid)+query(k->ch[1],k->mid+1,r);
}
void single_change(Segment *&k,int t,int v)
{
    if(k->l==k->r)
    {
        k->upval+=v;
        return;
    }
    if(t<=k->mid) single_change(k->ch[0],t,v);
    else single_change(k->ch[1],t,v);
    pushup(k);
}
int Main()
{
    Read(n);
    Read(m);
    for(int i=1;i<=30;++i) IsLucky[Lucky[i]]=1;
    for(int i=1;i<=n;++i) scanf("%d",&dis[i]);
    Segment *root=new Segment;
    build(root,1,n);
    char str[10];
    for(int x,y,z;m--;)
    {
        scanf("%s",str+1);
        if(str[1]=='c')
        {
            Read(x);
            Read(y);
            printf("%d
",query(root,x,y));
        }
        else 
        {
            Read(x);
            Read(y);
            Read(z);
            for(int i=x;i<=y;++i)
            {
                int v=0;
                if(IsLucky[dis[i]]) --v; 
                dis[i]+=z;
                if(IsLucky[dis[i]]) ++v;
                if(v) single_change(root,i,v);
            }
        }
    }
    return 0;
}
int sb=Main();
int main(int argc,char *argv[]){;}

 树状数组

#include <ctype.h>
#include <cstdio>
const int N = 1e6+5;
bool IsLucky[N];
int tag[N],dis[N],n,m,Lucky[50]={0,4,7,44,47,74,77,444,447,474,477,744,747,774,777,4444,4447,4474,4477,4744,4747,4774,4777,7444,7447,7474,7477,7744,7747,7774,7777};
inline void Read(int &x)
{
    bool f=0;register char ch=getchar();
    for(x=0;!isdigit(ch);ch=getchar()) if(ch=='-') f=1;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
    x=f?-x:x;
}
inline int lowbit(int x) {return x&(-x);}
inline void modify(int x,int y)
{
    for(;x<=n;x+=lowbit(x)) tag[x]+=y;
}
inline int ask(int x)
{
    int sum=0;
    for(;x;x-=lowbit(x)) sum+=tag[x];
    return sum;
}
int main()
{
    Read(n);Read(m);
    for(int i=1;i<=30;++i) IsLucky[Lucky[i]]=1;
    for(int i=1;i<=n;++i) {Read(dis[i]);if(IsLucky[dis[i]]) modify(i,1);} 
    char str[10];
    for(int x,y,z;m--;)
    {
        scanf("%s",str+1);
        if(str[1]=='c')
        {
            Read(x);
            Read(y);
            printf("%d
",ask(y)-ask(x-1));
        }
        else 
        {
            Read(x);
            Read(y);
            Read(z);
            for(int i=x;i<=y;++i)
            {
                int v=0;
                if(IsLucky[dis[i]]) --v;
                dis[i]+=z;
                if(IsLucky[dis[i]]) ++v;
                if(v) modify(i,v);
            }
        }
    }
    return 0;
}