POJ 3468 A Simple Problem with Integers（线段树）

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 117447 Accepted: 36521Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

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

Sample Input

10 51 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

线段树功能:update:成段增减 query:区间求和

代码：

#include <cstdio>  #include <algorithm>  using namespace std;     #define lson l , m , rt << 1  #define rson m + 1 , r , rt << 1 | 1  #define LL long long  const int maxn = 111111;  LL add[maxn<<2];  LL sum[maxn<<2];  void PushUp(int rt) {      sum[rt] = sum[rt<<1] + sum[rt<<1|1];  }  void PushDown(int rt,int m) {      if (add[rt]) {          add[rt<<1] += add[rt];          add[rt<<1|1] += add[rt];          sum[rt<<1] += add[rt] * (m - (m >> 1));          sum[rt<<1|1] += add[rt] * (m >> 1);          add[rt] = 0;      }  }  void build(int l,int r,int rt) {      add[rt] = 0;      if (l == r) {          scanf("%lld",&sum[rt]);          return ;      }      int m = (l + r) >> 1;      build(lson);      build(rson);      PushUp(rt);  }  void update(int L,int R,int c,int l,int r,int rt) {      if (L <= l && r <= R) {          add[rt] += c;          sum[rt] += (LL)c * (r - l + 1);          return ;      }      PushDown(rt , r - l + 1);      int m = (l + r) >> 1;      if (L <= m) update(L , R , c , lson);      if (m < R) update(L , R , c , rson);      PushUp(rt);  }  LL query(int L,int R,int l,int r,int rt) {      if (L <= l && r <= R) {          return sum[rt];      }      PushDown(rt , r - l + 1);      int m = (l + r) >> 1;      LL ret = 0;      if (L <= m) ret += query(L , R , lson);      if (m < R) ret += query(L , R , rson);      return ret;  }  int main() {      int N , Q;      scanf("%d%d",&N,&Q);      build(1 , N , 1);      while (Q --) {          char op[2];          int a , b , c;          scanf("%s",op);          if (op[0] == 'Q') {              scanf("%d%d",&a,&b);              printf("%lld\n",query(a , b , 1 , N , 1));          } else {              scanf("%d%d%d",&a,&b,&c);              update(a , b , c , 1 , N , 1);          }      }      return 0;  }