POJ 3468 A Simple Problem with Integers

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.

题目大意：

有N个数分别为A1 - An， 下面给出这样几种操作：  1.在给定区间内将每个数字都加上某一特定值   2.查询给定区间的数字和。实现上述操作。

解题思路：

根据题目意思可以看出这道题是线段树的区间增值和区间查询和的模板题， 只要根据题目描述更改模板即可。

代码：

#include <iostream>#include <fstream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <climits>#include <cmath>#include <queue>#include <stack>#include <deque>#include <vector>#include <map>#include <set>#include <iomanip>using namespace std;//definitionconst int Max = 110010;int tree[Max];long long st[Max * 4], add[Max * 4], change[Max * 4];void Pushup(int o){    // update parent node & return    st[o] = st[o << 1] + st[(o << 1) + 1];}// deliver node to left and right childvoid Pushdown(int o, int len){    if (add[o]){        // add update parent to child        add[o << 1] += add[o];        add[(o << 1) + 1] += add[o];        // update node        st[o << 1] += (add[o] * (len - (len >> 1)));        st[(o << 1) + 1] += add[o] * (len >> 1);        add[o] = 0;    }}void Build(int o, int l, int r){     add[o] = 0;     if (l == r){        st[o] = tree[l];        return;     }     int mid = (l + r) >> 1;     Build(o << 1, l, mid);     Build((o << 1) + 1, mid + 1, r);     // perform the sum     Pushup(o);}/// section add & changinglong long Section_query(int o, int l, int r, int ql, int qr){   if (ql <= l && qr >= r)       return st[o];   Pushdown(o, r - l + 1);   int mid = (l + r) >> 1;   long long ans = 0;   if (ql <= mid) ans += Section_query(o << 1, l, mid, ql, qr);   if (qr > mid) ans += Section_query((o << 1) + 1, mid + 1, r, ql, qr);   return ans;}// last parameter, child, terminal for section, value for addvoid Section_update(int o, int l, int r, int tl, int tr, int value){    // in this section or not    if (tl <= l && tr >= r){        add[o] += value;        st[o] += (long long)value * (r - l + 1);        return;    }    Pushdown(o, r - l + 1);    int mid = (r + l) >> 1;    // find the update section    if (tl <= mid) Section_update(o << 1, l, mid, tl, tr, value);    if (tr > mid) Section_update((o << 1) + 1, mid + 1, r, tl, tr, value);    // update parent node    Pushup(o);}int main(){    // definition    int n, m;    scanf("%d%d", &n ,&m);    for (int i = 1; i <= n; ++i){        scanf ("%d", &tree[i]);    }    // test    Build(1, 1, n);    while(m--){        char c[5];        scanf("%s",c);        if (c[0] == 'Q'){            int left, right;            scanf("%d %d", &left, &right);            printf("%I64d\n", Section_query(1, 1, n, left, right));        }        else{            int left, right, val;            scanf("%d %d %d", &left, &right, &val);            Section_update(1, 1, n, left, right, val);        }    }    return 0;}