POJ 3468 A Simple Problem with Integers 线段树 区间修改
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A Simple Problem with Integers
Description You have N integers, A1, A2, ... , AN. 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. 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
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点击打开题目链接
想要写点什么，又不知怎么写。（ps：刚刚入门）
（蓝桥，省赛，天梯。。。一堆的比赛就要将自己的小身板压垮了，坚持！想想两个月后就要加入了程序猿的队伍，期待又胆怯。
越努力，越幸运---致默默为祖国IT事业做出自己贡献的网络搬运工）
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long LL;
#define Lson 2 * o, L, M //左儿子 宏定义 会有意想不到的好处
#define Rson 2 * o + 1, M + 1, R //右儿子
#define lo 2 * o
#define ro 2 * o + 1
const int maxn = 4 * 100000 + 10;
int ql, qr, v;
long long sum[maxn], add[maxn];
//向上传递，由 o 的左右子树更新 o
void pushUp(int o)
{
sum[o] = sum[lo] + sum[ro];
}
//向下传递，维护节点o
void pushDown(int o, int m)
{
if (add[o]) //本节点有标记才传递
{
add[lo] += add[o]; //左子树点增加
add[ro] += add[o]; //右子树点增加
sum[lo] += add[o] * (m - (m / 2)); //左子树区间段增加
sum[ro] += add[o] * (m / 2); //右子树区间段增加
add[o] = 0; //清除本节点标记
}
}
//建树
void Build(int o, int L, int R)
{
add[o] = 0; //建树时初始节点都未标记
if (L == R) scanf("%lld", &sum[o]); //叶子节点
else
{
int M = (L + R) / 2;
Build(Lson); //建立左子树
Build(Rson); //建立右子树
pushUp(o); //更新本节点
}
}
//更新区间 [ql,qr] 内的值加上 v
void Update(int o, int L, int R)
{
if (ql <= L && qr >= R) //递归边界
{
add[o] += v; //累加边界的add值
sum[o] += v * (R - L + 1); //计算区间和
}
else
{
pushDown(o, R - L + 1);
int M = (L + R) / 2;
if (ql <= M) Update(Lson); //先递归更新左子树或者右子树
if (qr > M) Update(Rson);
pushUp(o); //更新本节点
}
}
//查询区间[ql,qr] 的和
LL Query(int o, int L, int R)
{
if (ql <= L && qr >= R) //递归边界
{
return sum[o];
}
else
{
pushDown(o, R - L + 1); //用边界区间的附加信息更新答案
int M = (L + R) / 2;
LL ans = 0; //递归统计，累加参数add
if (ql <= M) ans += Query(Lson);
if (qr > M) ans += Query(Rson);
return ans;
}
}
int main()
{
int n, q;
while (~scanf("%d%d", &n, &q))
{
Build(1, 1, n);
char ch[5];
for (int i = 1; i <= q; i++)
{
scanf("%s", ch);
if (ch[0] == 'Q')
{
scanf("%d%d", &ql, &qr);
printf("%lld\n", Query(1, 1, n));
}
else
{
scanf("%d%d%d", &ql, &qr, &v);
Update(1, 1, n);
}
}
}
return 0;
}