http://www.elijahqi.win/archives/1838
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
題意 要求我們求一段區間內 幸運數字有多少個 對於幸運數字的定義是：只能包含十進制下的4&7而且題目保證這個他加的數最終不會超過1e4那麼我們考慮這題是否和有一個開根號的那題相似呢 大概leoly說很像呢 所以做法就是 我對於線段樹上每個點我去儲存d[i]表示我的值到距離我最近的那個幸運數字的差是多少 然後維護一個這些值的最小值 維護一下這個最小值最早出現在哪裏 再維護一下我最小值一共有幾個 然後修改的時候我相當於是區間減法 然後如果這個整個所有1~n中出現了負數 那麼我就定位到第一個負數出現在哪裏把他改成距離下一個幸運數字的大小 然後重新更新上去 最後直到我1~n中不再有<0的數出現即可 詢問的時候只需要詢問我最小值爲0的數有多少個即可
#include<queue>
#include<cstdio>
#include<algorithm>
#define N 110000
#define inf 0x3f3f3f3f
#define pa pair<int,int>
using namespace std;
inline int read(){
int x=0;char ch=getchar();
while(ch<'0'||ch>'9') ch=getchar();
while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
return x;
}
struct node{
int left,right,min,mp,mn,v,lazy;
}tree[N<<2];
int mp[]={0,4,7,44,47,74,77,444,447,474,744,477,747,774,777,4444,4447,4474,4744,7444,4477,4747,7447,4774,7474,7744,4777,7477,7747,7774,7777,999999999};
int n,m,a[N],num,root;char s[30];
inline void update(int x){
int l=tree[x].left,r=tree[x].right;
if (tree[l].min==tree[r].min){
tree[x].mn=tree[l].mn+tree[r].mn;
tree[x].min=tree[l].min;tree[x].mp=min(tree[l].mp,tree[r].mp);
}
if (tree[l].min<tree[r].min){
tree[x].min=tree[l].min;tree[x].mn=tree[l].mn;tree[x].mp=tree[l].mp;
}
if (tree[r].min<tree[l].min){
tree[x].min=tree[r].min;tree[x].mn=tree[r].mn;tree[x].mp=tree[r].mp;
}
}
inline void pushdown(int x){
if (!tree[x].lazy) return;
int l=tree[x].left,r=tree[x].right;
tree[l].lazy+=tree[x].lazy;tree[r].lazy+=tree[x].lazy;
tree[l].min-=tree[x].lazy;tree[r].min-=tree[x].lazy;
tree[x].lazy=0;
}
void build(int &x,int l,int r){
x=++num;
if (l==r){
int pos=lower_bound(mp+1,mp+32,a[l])-mp;
tree[x].v=mp[pos];tree[x].min=mp[pos]-a[l];tree[x].mp=l;tree[x].mn=1;return;
}
int mid=l+r>>1;
build(tree[x].left,l,mid);build(tree[x].right,mid+1,r);update(x);
}
inline void change(int x,int l,int r,int p){
if(l==r){
int now=tree[x].v-tree[x].min,pos=lower_bound(mp+1,mp+32,now)-mp;
tree[x].v=mp[pos];tree[x].min=mp[pos]-now;return;
}
int mid=l+r>>1;pushdown(x);
if(p<=mid) change(tree[x].left,l,mid,p);else change(tree[x].right,mid+1,r,p);update(x);
}
inline pa update1(pa p1,pa p2){
int fi,se;
if(p1.first<p2.first) fi=p1.first,se=p1.second;
if(p1.first==p2.first) fi=p1.first,se=p1.second+p2.second;
if(p1.first>p2.first) fi=p2.first,se=p2.second;return make_pair(fi,se);
}
inline pa qr(int x,int l,int r,int l1,int r1){
if(l1<=l&&r1>=r){pa pp;pp=make_pair(tree[x].min,tree[x].mn);return pp;}
int mid=l+r>>1;pa tmp;tmp=make_pair(inf,99);pushdown(x);
if(l1<=mid) tmp=update1(tmp,qr(tree[x].left,l,mid,l1,r1));
if(r1>mid) tmp=update1(tmp,qr(tree[x].right,mid+1,r,l1,r1));return tmp;
}
inline void insert1(int x,int l,int r,int l1,int r1,int v){
if(l1<=l&&r1>=r){tree[x].lazy+=v;tree[x].min-=v;return;}
int mid=l+r>>1;pushdown(x);
if(l1<=mid) insert1(tree[x].left,l,mid,l1,r1,v);
if(r1>mid) insert1(tree[x].right,mid+1,r,l1,r1,v);update(x);
}
int main(){
freopen("cf.in","r",stdin);
n=read();m=read();sort(mp+1,mp+32);
//for (int i=1;i<=31;++i) printf("%d\n",mp[i]);
for (int i=1;i<=n;++i) a[i]=read();build(root,1,n);
for (int i=1;i<=m;++i){
scanf("%s",s+1);if (s[1]=='c'){
int l=read(),r=read();pa pp=qr(root,1,n,l,r);
if(pp.first==0) printf("%d\n",pp.second);else printf("0\n");
} else{
int l=read(),r=read(),v=read();insert1(root,1,n,l,r,v);
while(tree[root].min<0) change(root,1,n,tree[root].mp);
}
}
return 0;
}