分块的学习

模板题地址

(Juan Feng) : 一定可以卡过去的！

• 数列长度是(N)，把一列数分成(sqrt{N})个块，基于分治策略完成。
• (Sum[i]) 第(i)块中数的和
• (Tag[i])第(i)块中数字被整体修改的值
• (A[i])暴力维护即可。

#include <string.h>
#include <stdio.h>
#include <math.h>
#define GC getchar()
#define Clean(X,K) memset(X,K,sizeof(X))
const int Maxn = 100005 , MaxR = 320 ;
int N , M  , Rt , Belong[Maxn];
long long A[Maxn] ;
long long Qread () {
    long long X = 0 ;
    char C = GC ;
    while (C > '9' || C < '0') C = GC ;
    while (C >='0' && C <='9') {
        X = X * 10 + C - '0' ;
        C = GC ;
    }
    return X ;
}
struct Block {
    int Left , Right ;
    long long Tag , Sum ;
};
Block B[MaxR] ;
long long Ask (int L , int R) {
    int St = Belong[L] , Ed = Belong[R] ;
    long long Ans = 0 ;
    if (St == Ed) {
        for (int i = L ; i <= R ; ++ i) Ans += A[i] ;
        Ans += B[St].Tag * (R - L + 1) ;
        return Ans ;
    }
    for (int i = L ; i <= B[St].Right ; ++ i) Ans += A[i] ;
    Ans += B[St].Tag * (B[St].Right - L + 1) ;
    for (int i = B[Ed].Left ; i <= R ; ++ i) Ans += A[i] ;
    Ans += B[Ed].Tag * (R - B[Ed].Left + 1) ;
    for (int i = St + 1 ; i < Ed ; ++ i) Ans += B[i].Sum ;
    return Ans ;
}
void Add (int L , int R , long long int K) {
    int St = Belong[L] , Ed = Belong[R] ;
    if (St == Ed) {
        B[St].Sum += (R - L + 1) * K ;
        for (int i = L ; i <= R ; ++ i) A[i] += K ;
        return ;
    }
    B[St].Sum += (B[St].Right - L  + 1) * K ;
    B[Ed].Sum += (R - B[Ed].Left  + 1) * K ;
    for (int i = L ; i <= B[St].Right ; ++ i) A[i] += K ;
    for (int i = B[Ed].Left ; i <= R ; ++ i) A[i] += K ;
    for (int i = St+ 1 ; i < Ed ; ++ i) {
        B[i].Sum += Rt * K ;
        B[i].Tag += K ;
    }
}
int main () {
//	freopen ("P3372.txt" , "r" , stdin) ;
//	freopen ("my.out" , "w" , stdout) ;
    N = Qread () , M = Qread () ;
    Rt = sqrt ((double)N) + 1 , Clean(Belong , 0) , Clean(B , 0);
    for (int i = 1 ; i <= N; ++ i) A[i] = Qread () ;
    for (int i = 0 ; i <= Rt; ++ i) {
        B[i].Left = i * Rt + 1, B[i].Right = B[i].Left + Rt - 1 ;
        for (int j = B[i].Left ; j <= B[i].Right&&j <= N ; ++ j) Belong[j] = i , B[i].Sum += A[j];
    }
    for (int i = 1 ; i <= M ; ++ i) {
        int Q = Qread () , L = Qread () , R = Qread ();
        if (Q == 2) printf ("%lld
" , Ask (L , R)) ;
        else {
            long long K = Qread () ;
            Add (L , R , K) ;
        }
    }
    fclose (stdin) , fclose (stdout) ;
    return 0 ;
}