POJ 3468--a simple problem with integers —————— "segment tree interval update, interval query"

A simple problem with integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 86780		Accepted: 26950
Case Time Limit: 2000MS

Description

You have N integers, a1, a2, ..., an. You need to deal with both kinds of operations. One type of operation is to add some given number to each number in a given interval. The other are to ask for the sum of numbers in a given interval.

Input

The first line contains the numbers N and Q. 1 ≤ N,Q ≤100000.
The second line contains N numbers, the initial values of a1, a2, ..., an. -1000000000≤ Ai ≤1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding C to each of AA, aa+1, ..., ab. -10000≤ C ≤10000.
"Q a b" means querying the sum of aa, aa+1, ..., Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ monthly--2007.11.25, Yang Yi topic: give you n number, m times ask. Queries include the addition of C for each value in the interval and all values within the query interval. Problem-solving ideas: delay markers. Ensure that the current value is correct. Lazy cannot be emptied and should accumulate because the last lazy may not have been pushed down.

#include <stdio.h> #include <vector> #include <queue> #include <string.h> #include < Algorithm>using namespace Std;typedef long long ll;const int MAXN = 1e6+20;const int INF = 0x3f3f3f3f;const int mod = 1 e9+7; #define MID (L+r)/2#define Lson rt*2,l,mid#define rson rt*2+1,mid+1,rstruct segtree{LL val, lazy;} segs[maxn*4];void pushup (int rt) {Segs[rt].val = Segs[rt*2+1].val + segs[rt*2].val;}        void pushdown (int rt,int l,int R) {if (Segs[rt].lazy) {segs[rt*2].val + = segs[rt].lazy* (mid-l + 1);//Guaranteed Current Correctness        Segs[rt*2+1].val + = segs[rt].lazy* (R-mid);   Segs[rt*2].lazy + = Segs[rt].lazy;        Cumulative Segs[rt*2+1].lazy + = Segs[rt].lazy;    Segs[rt].lazy = 0;    }}void buildtree (int rt,int l,int R) {segs[rt].val = 0;    Segs[rt].lazy = 0;        if (L = = R) {scanf ("%lld", &segs[rt].val);    return;    } buildtree (Lson);    Buildtree (Rson); Pushup (RT);} void Update (int rt,int l,int r,int _idx,int _val) {//if (L = = R && L= = _idx) {//Segs[rt].val + = _val;//return;//}//if (_idx <= mid)//Update (Lson,_idx,_val); else//Update (rson,_idx,_val);//pushup (RT);//}ll query (int rt,int l,int r,int l_ran,int R_ran) {if (l_r    An <= l&&r <= R_ran) {return segs[rt].val;    } pushdown (Rt,l,r);    LL ret = 0;    if (L_ran <= mid) {ret + = query (Lson,l_ran,r_ran);    } if (R_ran > mid) {ret + = query (Rson,l_ran,r_ran);    } pushup (RT); return ret;} void Update (int rt,int l,int r,int l_ran,int r_ran,ll chg) {if (L_ran <= l&&r <= r_ran) {SEGS[RT].V        Al + = chg* (r-l+1);        Segs[rt].lazy + = Chg;    return;    } pushdown (Rt,l,r);    if (L_ran <= mid) Update (LSON,L_RAN,R_RAN,CHG);    if (R_ran > Mid) Update (RSON,L_RAN,R_RAN,CHG); Pushup (RT);}    int main () {int n,m;        while (scanf ("%d%d", &n,&m)!=eof) {buildtree (1,1,n);        int l,r;        LL C;        Char s[12];for (int i = 1; I <= m; i++) {scanf ("%s", s);                if (s[0]== ' Q ') {scanf ("%d%d", &l,&r);                LL ans = query (1,1,n,l,r);            printf ("%lld\n", ans);                }else{scanf ("%d%d%lld", &l,&r,&c);            Update (1,1,N,L,R,C); }}} return 0;}

　　

POJ 3468--a simple problem with integers —————— "segment tree interval update, interval query"