HDU 4288 line segment tree sum of medians like lazysales girl coder

Question:

That is, the number is inserted continuously, and the number is pushed from small to large. Calculate the sum of the numbers 3, 8, 13.

Solution:

Line Segment tree. The coordinates of the Line Segment tree indicate the largest number of all given numbers (0 is not yet,> 0 is the largest number of I already inserted ). Then ans [] [I] indicates the sum of the Number % 5 of the current interval (subscripts I (I) (which may be a bit unclear ..) Yes, yes. The last query is O (1.

Question preparation process:

Finally, there was a question about the idea of an online competition... At that time, the idea was somewhat biased. During the competition, the node meaning of my line segment tree is correct, that is, I actually represent all the entire intervals... As a result, I don't know how to update the update after me .... In addition, this is my upcoming question. I have to speed up my question preparation. If the competition is like this, I really regret it.

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # define maxn 111111 # define lson l, m, RT <1 # define rson m + 1, R, RT <1 | 1 # define havem int M = (L + r)> 1 using namespace STD; int sum [maxn <2], n, x [maxn], flag, add [maxn], xsub ;__ int64 ans [maxn <2] [5]; char op [maxn] [5]; void build (int l, int R, int RT) {for (INT I = 0; I <5; I ++) ans [RT] [I] = 0; sum [RT] = 0; If (L = r) Return; havem; build (lson); Build (rson);} void push_up (int rt) {for (INT I = 0; I <5; I ++) ans [RT] [I] = ans [RT <1] [I] + ans [RT <1 | 1] [(I-sum [RT <1]) % 5 + 5) % 5]; // it must be the remainder plus 5 and then the remainder} void Update (INT POs, int L, int R, int RT) {sum [RT] + = 2 * flag-1; if (L = r) {ans [RT] [0] = x [POS] * flag; return ;} int M = (L + r)> 1; if (Pos <= m) Update (Pos, lson); else Update (Pos, rson); push_up (RT );} int Main () {While (~ Scanf ("% d", & N) {xsub = 0; For (INT I = 0; I <n; I ++) {scanf ("% s ", OP [I]); If (OP [I] [0] = 'A' | op [I] [0] = 'D ') scanf ("% d", & add [I]), X [xsub ++] = add [I];} Sort (x, x + xsub ); xsub = unique (x, x + xsub)-X; build (0, xsub, 1); // start from 1 for (INT I = 0; I <N; I ++) {If (OP [I] [0]! ='S ') {int low = lower_bound (x, x + xsub, add [I])-X; If (OP [I] [0] = 'A ') flag = 1; else flag = 0; Update (low, 0, xsub, 1);} else printf ("% i64d \ n ", ans [1] [2]) ;}} return 0 ;}

The code of the competition...

/*Pro: 0Sol:date:*/#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <set>#include <vector>#define lson l , m , rt << 1#define rson m + 1, r , rt << 1 | 1#define maxn 111111 //using namespace std;int X[maxn],n,add[maxn],xsub;__int64 p[maxn << 2][6];char op[maxn][10];short pp[maxn];void build(int l, int r, int rt){    for(int i = 0; i < 5; i ++)        p[rt][i] = 0;    if(l == r) return ;    int m = (l + r) >> 1;    build(lson); build(rson);}void push_up(int rt){    for(int i = 0; i < 5; i ++)        p[rt][i] = p[rt << 1][i] + p[rt << 1 | 1][i];}void update(int pos, short ii, int c, int l,int r, int rt){//    if(l == r) {        p[rt][ii] += c * X[pos];        return ;    }    int m = (l + r) >> 1;    if(pos <= m) update(pos,ii,c,lson);    else update(pos,ii,c,rson);    push_up(rt);}int main(){    while(~scanf("%d",&n)){        xsub = 0;        for(int i = 0; i < n; i ++){            scanf("%s",&op[i]);            if(op[i][0] != 's')                scanf("%d",&add[i]),                X[xsub ++] = add[i];        }        sort(X, X + xsub);        int m = 1;        for(int i = 1; i <xsub; i ++){            if(X[i] != X[i - 1]) X[m ++] = X[i];        }        build(0, m, 1);        int x = 0;        for(int i = 0; i < n; i ++){            if(op[i][0] == 'a'){                int ll = lower_bound(X,X + m,add[i]) - X;                x ++, update(ll, x % 5, 1, 0 , m , 1);//???                pp[ll] = x % 5;            }            else if(op[i][0] == 'd'){                int ll = lower_bound(X,X + m,add[i]) - X;                x -- , update(ll, pp[ll], -1, 0,m, 1);            }            else{                if(3 > x)                    printf("0\n");                else{                    printf("%I64d  !!!  \n",p[1][3]);                }            }        }    }return 0;}