POJ3468,A Simple Problem with Integers,线段树
题目链接：
POJ3468
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 77541 | Accepted: 23874 | |
| Case Time Limit: 2000MS | ||
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.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
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.
题意：
输入N个数，有Q个操作，Q a b 表示查询区间[a,b]内所有数的和；C a b c 表示区间[a,b]内所有数都加上c
题解：
线段树区间更新，需要注意数的范围，得用long long
//bei
#include
#include
#include
#include
using namespace std;
#define lson l,m,rt<<1 // 树的左儿子
#define rson m+1,r,rt<<1|1 // 树的右儿子
#define Maxn 1111111
typedef long long ll;
ll sum[Maxn],col[Maxn]; // 数组得开大一点，不然会RE的
void PushUp(ll rt) // 向上更新
{
sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void build(ll l,ll r,ll rt) // 建立线段树
{
if (l == r)
{
scanf("%lld",&sum[rt]);
return ;
}
ll m = (l+r) >> 1;
build(lson);
build(rson);
PushUp(rt);
}
void PushDown(ll rt, ll m) // 线段树延迟标记
{
if (col[rt]) // 向下标记的时候，col[rt]的值可能是负数，所以这里如果写 col[rt] > 0 就gg了
{
// 下面的四条语句都是 += ，不是 = ，因为题目说的是区间[a,b]内所有数加上c
col[rt<<1] += col[rt]; // 如果不是很理解 col 为什么也是 += 的话，可以举个例子加以理解
col[rt<<1|1] += col[rt];
sum[rt<<1] += (m-(m>>1)) * col[rt];
sum[rt<<1|1] += (m>>1) * col[rt];
col[rt] = 0;
}
}
void update(ll L, ll R, ll c, ll l, ll r, ll rt) // 区间更新
{
if (L <= l && r <= R)
{
// 下面两条语句也必须是 +=
col[rt] += c;
sum[rt] += c * (r-l+1);
return ;
}
PushDown(rt,r-l+1);
ll m = (l+r) >> 1;
if (L <= m)
{
update(L,R,c,lson);
}
if (R > m)
{
update(L,R,c,rson);
}
PushUp(rt);
}
ll query(ll L, ll R, ll l, ll r, ll rt) // 区间查询
{
if (L <= l && r <= R)
{
return sum[rt];
}
PushDown(rt,r-l+1); // 查询的时候也必须得向下标记，这条语句不能少！不明白的话，可以举个例子加以理解
ll m = (l+r) >> 1;
ll ans = 0;
if (L <= m)
ans += query(L,R,lson);
if (R > m)
ans += query(L,R,rson);
return ans;
}
int main()
{
ll N,Q,a,b,c;
char ch[3];
while (~scanf("%lld%lld",&N,&Q))
{
memset(sum,0,sizeof(sum));
memset(col,0,sizeof(col));
build(1,N,1);
while (Q--)
{
scanf("%s",ch);
if (ch[0] == 'C')
{
scanf("%lld%lld%lld",&a,&b,&c);
update(a,b,c,1,N,1);
}
else if (ch[0] == 'Q')
{
scanf("%lld%lld",&a,&b);
printf("%lld\n",query(a,b,1,N,1));
}
}
}
return 0;
}