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\n",_sum);

        }

        else

        {

            scanf("%I64d%I64d%I64d",&qL,&qR,&v);

            tree.update(1,1,n);

        }

    }

   // system("pause");

  return 0;

}