Bzoj3343 's Magic (permission title)

Bzoj Portal (permission)

Description

The leader has recently learned a magical magic that can make a person grow taller. So he was ready to demonstrate to each hero of the Xmyz Information Group. So the N heroes gathered together again, this time they were in a row numbered 1, 2 、......、 N.
Each person's height begins with a positive integer that does not exceed 1000. The Master's magic can add an integer w to each hero's height in the closed interval [L, R] (1≤l≤r≤n). (although L=r does not conform to the written specification of the interval, we can think of it as a separate increase in the height of the hero of the L (R))
CYZ, light elder brother and Zjq and other people do not worship the evil of the Lord, so they sometimes ask the WD closed interval [L, R] How many heroes height is greater than or equal to C, to verify that the master's magic is really effective.
WD was lazy, so he gave you the task of answering.

Input

1th Act two integers n, Q. Q is the sum of the number of questions and the number of spells cast.
The 2nd line has n positive integers, and the number of I represents the height of the hero of the first I.
3rd to Q+2 Line has one action per line:
(1) If the first letter is "M", then there are three numbers L, R, W. Represents the height of all heroes in the closed interval [L, R] plus W.
(2) If the first letter is "A", then there are three numbers L, R, C. Ask how many heroes in the closed interval [L, R] are taller than or equal to C.

Output

For each "A" query, the output line contains only one integer, representing the number of heroes in the closed interval [L, R] that are greater than or equal to C.

Sample Input

5 3

1 2 3) 4 5

A 1 5 4

M 3 5 1

A 1 5 4

Sample Output

2

3

HINT

"Input and Output sample description"

The original 5 heroes were 1, 2, 3, 4, 5, and at this time [1, 5] There were 2 heroes taller than or equal to 4. The master cast to 1, 2, 4, 5, 6, at this time [1, 5] There are 3 heroes taller than or equal to 4.

"Data Range"

For 30% of the data, n≤1000,q≤1000.

For 100% of the data, n≤1000000,q≤3000,1≤w≤1000,1≤c≤1,000,000,000.
Source

The first piece of life in the problem, feel good without words ah. The speed is still too slow, open the O2 before

CODE:

#include <cmath> #include <cstdio> #include <algorithm> using namespace std;
const int N=1E6+10;
    struct Block {int l,r,num,plus;
    int a[1010];
    inline void Keep () {sort (a+1,a+num+1);
        } inline int find (int n) {int l=1,r=num,mid;
        N-=plus;
            while (L<r) {mid= (l+r) >>1;
            if (a[mid]<n) l=mid+1;
        else R=mid;
    } return L;
}}a[1010];
int s[n];
int belong[n];
int n,q,x,y,z,ans,block,num;
char c;
    inline void read (int &n) {N=0;char C=getchar (); while (c< ' 0 ' | |
    C> ' 9 ') C=getchar ();
while (c>= ' 0 ' &&c<= ' 9 ') N=n*10+c-48,c=getchar ();
    } inline void Build () {block=sqrt (n);
    Num=n/block;if (N%block) num++;
    for (int i=1;i<=num;i++) a[i].l= (i-1) *block+1,a[i].r=i*block,a[i].num=block;
    a[num].r=n;a[num].num=a[num].r-a[num].l+1;

    for (int i=1;i<=n;i++) belong[i]= (i-1)/block+1; for (int i=1;i<=num;i++) {for (int j=a[i].l;j<=a[i].r;j++) A[I].A[J-A[I].L+1]=S[J];
    A[i].keep (); }} inline void Add (int x,int y,int z) {if (Belong[x]==belong[y]) {for (int i=x;i<=y;i++) s[i
        ]+=z;
        for (int i=a[belong[x]].l;i<=a[belong[x]].r;i++) a[belong[x]].a[i-a[belong[x]].l+1]=s[i];
        A[belong[x]].keep ();
    Return
    } for (int i=x;i<=a[belong[x]].r;i++) s[i]+=z;
    for (int i=a[belong[x]].l;i<=a[belong[x]].r;i++) a[belong[x]].a[i-a[belong[x]].l+1]=s[i];
    A[belong[x]].keep ();
    for (int i=belong[x]+1;i<belong[y];i++) a[i].plus+=z;
    for (int i=a[belong[y]].l;i<=y;i++) s[i]+=z;
    for (int i=a[belong[y]].l;i<=a[belong[y]].r;i++) a[belong[y]].a[i-a[belong[y]].l+1]=s[i];
A[belong[y]].keep ();
    } inline int Ask (int x,int y,int z) {int ans=0; if (Belong[x]==belong[y]) {for (int i=x;i<=y;i++) if (s[i]+a[belong[x]].plus>=z) ans++;
    return ans;
    } for (int i=x;i<=a[belong[x]].r;i++) if (s[i]+a[belong[x]].plus>=z) ans++;
        for (int i=belong[x]+1;i<belong[y];i++) {int p=a[i].find (z);
    if (a[i].a[p]+a[i].plus>=z) ans+=a[i].num-p+1;
    } for (int i=a[belong[y]].l;i<=y;i++) if (s[i]+a[belong[y]].plus>=z) ans++;
return ans;
    } int main () {read (n), read (q);
    for (int i=1;i<=n;i++) read (s[i]);
    Build ();
        while (q--) {C=getchar ();
        while (c!= ' A ' &&c!= ' M ') C=getchar ();
        Read (x), read (y), read (z);
        if (c== ' A ') printf ("%d\n", ask (x, Y, z));
    else Add (x, y, z);
} return 0; }