Hdoj 4267-line segment tree

2012 Changchun division network competition... Question

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4267

01 there are two operations for Array ai with an initial value
1. Add C to the value of each AI that satisfies (I-A) % K = 0 in the range [a, B]
2. Ask about the AI value.

The idea of forming a line segment tree is obvious, because k <= 10, the value is very small, so the Maintenance Interval Values of nodes can be 1-10, a total of 55 cases
Add [10] [10] directly to sb during the competition... keep MLE all the time, and change it to dynamic allocation. After TLE, Nima knows that adding [55] is just right ....

Don't talk about anything, postCode... Depressed ..

# Include <stdio. h> # include <string. h >#include <algorithm> # include <cmath> # include <iostream> using namespace STD; const int n = 50002; int n, m; struct node {bool flag; int L, R; int val; int add [55]; void reset () {memset (ADD, 0, sizeof (ADD); flag = false ;}} A [n * 4]; int B [N], KK, CC; void build (INT L, int R, int p) {A [p]. L = L; A [p]. R = r; A [p]. reset (); If (L = r) {A [p]. val = B [l]; return;} int mid = (L + r)> 1; build (L, mi D, P * 2); Build (Mid + 1, R, p * 2 + 1);} inline int ijtonum (int I, Int J) {return (I * (I + 1)/2 + J;} inline void down (INT p) {if (a [p]. flag = true) {int I, j; for (I = 0; I <10; I ++) {int II = I + 1; for (j = 0; j <= I; j ++) {if (a [p]. add [ijtonum (I, j)]! = 0) {A [p * 2]. add [ijtonum (I, j)] + = A [p]. add [ijtonum (I, j)]; A [p * 2]. flag = true; A [p * 2 + 1]. add [ijtonum (I, (II * 50001-(A [p * 2]. r-A [p * 2]. L + 1-j) % II)] + = A [p]. add [ijtonum (I, j)]; // note the influence of the Offset generated in the left half on the second half... That is, the use of j... A [p * 2 + 1]. flag = true ;}} A [p]. reset () ;}} void Update (int l, int R, int P, int KI) {if (a [p]. L = L & A [p]. R = r) {// A [p]. add [kk-1] [Ki] + = cc; A [p]. add [ijtonum (kk-1, KI)] + = cc; A [p]. flag = true; return;} // down (p); int mid = (a [p]. L + A [p]. r)> 1; if (r <= mid) Update (L, R, p * 2, KI); else if (L> mid) Update (L, R, p * 2 + 1, KI); else {Update (L, mid, p * 2, KI); Update (Mid + 1, R, p * 2 + 1, (kk * 50001-(mid-L + 1-ki) % KK); // you can create a [p] Here. R-A [p]. L + 1 ....}} Int query (int l, int R, int p) {if (a [p]. L = L & A [p]. R = r) {if (a [p]. flag = true) {for (INT I = 0; I <10; I ++) {// A [p]. val + = A [p]. add [I] [0]; A [p]. val + = A [p]. add [ijtonum (I, 0)]; // A [p]. add [I] [0] = 0; A [p]. add [ijtonum (I, 0)] = 0;} A [p]. flag = false;} return a [p]. val;} Down (p); int mid = (a [p]. L + A [p]. r)> 1; if (r <= mid) return query (L, R, p * 2); else if (L> mid) return query (L, R, p * 2 + 1);} int main () {int op, AA, BB; Wh Ile (scanf ("% d", & N )! = EOF) {for (INT I = 1; I <= N; I ++) {scanf ("% d", & B [I]);} build (1, n, 1); scanf ("% d", & M); // printf ("case # % d: \ n", cas1 ++); While (M --) {scanf ("% d", & OP); If (OP = 1) {scanf ("% d", & aa, & BB, & KK, & CC); Update (AA, BB, 1, 0);} else {scanf ("% d", & aa ); printf ("% d \ n", query (AA, AA, 1) ;}} return 0 ;}