poj 3468

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 71540		Accepted: 22049
Case 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 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

4
55
9
15

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

题目描述 ： 区间修改，区间查询

此题一直wa的原因是我的代码全部用long long 才能过，虽然不知道为什么。

// Fast Sequence Operations I
// Rujia Liu
// 输入格式：
// n m     数组范围是a[1]~a[n]，初始化为0。操作有m个
// 1 L R v 表示设a[L]+=v, a[L+1]+v, ..., a[R]+=v
// 2 L R   查询a[L]~a[R]的sum, min和max
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL  long long
using namespace std;

const int maxnode = 111111 << 2;

LL _sum;
LL qL,qR, v; //<span style="color:#ff0000;">_sum为全局变量</span>

struct IntervalTree {
  LL sumv[maxnode], addv[maxnode];

  // 维护信息
  void maintain(LL  o, LL L, LL  R) {
    int lc = o*2, rc = o*2+1;
    sumv[o]=0;
    if(R > L) {
      sumv[o] = sumv[lc] + sumv[rc];
    }
    if(addv[o]) {sumv[o] += addv[o] * (R-L+1); }
  }

  void update(LL o, LL  L, LL R) {
    int lc = o*2, rc = o*2+1;
    if(qL <= L && qR >= R) { // 递归边界
      addv[o] += v; // 累加边界的add值
    } else {
      int M = L + (R-L)/2;
      if(qL <= M) update(lc, L, M);
      if(qR > M) update(rc, M+1, R);
    }
    maintain(o, L, R); // 递归结束前重新计算本结点的附加信息
  }

  void query(LL  o, LL  L, LL  R, LL add) {
    if(qL <= L && qR >= R) { // 递归边界：用边界区间的附加信息更新答案
      _sum += sumv[o] + add * (R-L+1);
    } else { // 递归统计，累加参数add
      int M = L + (R-L)/2;
      if(qL <= M) query(o*2, L, M, add + addv[o]);
      if(qR > M) query(o*2+1, M+1, R, add + addv[o]);
    }
  }
};


IntervalTree tree;

int main() {

     int m,n;
    scanf("%d%d",&n,&m);
    memset(&tree,0,sizeof(tree));
    for(int i=1;i<=n;i++)
    {
      qL=i;qR=i;
      scanf("%I64d",&v);
      tree.update(1,1,n);
    }
    while(m--)
    {
        char s[5];
        int a,b,c;
        scanf("%s",s);
        if(s[0]=='Q')
        {
            scanf("%I64d%I64d",&qL,&qR);
            _sum=0;
            tree.query(1,1,n,0);
            printf("%I64d
",_sum);
        }
        else
        {
            scanf("%I64d%I64d%I64d",&qL,&qR,&v);
            tree.update(1,1,n);
        }
    }
   // system("pause");
  return 0;
}