POJ——3468A Simple Problem with Integers（区间更新）
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 78332 | Accepted: 24139 | |
| 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.
Source
POJ Monthly--2007.11.25, Yang Yi
这题WA了许久，最后发现读入数据时也应该是%I64d，而且用%LLd还不行
这题WA了许久，最后发现读入数据时也应该是%I64d，而且用%LLd还不行
/** 线段树：区间更新
*线段树每个节点存储当前区间的和
*标记数组存取当前增量，记得每次更新增量时要自加上之前的该区间增量，因为有可能之前该区间的增量并没有向下更新掉
*注意增量可能是负数
*输入输出用%I64d，%LLD会WA，应该和评测系统有关
*最好不用cin，cout，除非关掉同步
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#define rson mid + 1,r,rt << 1 |1
#define lson l,mid,rt << 1
#define LL long long
using namespace std;
const int N = 100005;
LL Tree[N << 2];
LL col[N << 2];
void PushUp(int rt)//向上更新和
{
Tree[rt] = Tree[rt << 1] + Tree[rt << 1 | 1];
}
void PushDown(int rt,int l)//向下更新增量
{
if(col[rt])
{
col[rt << 1] += col[rt];
col[rt << 1 | 1] += col[rt];
Tree[rt << 1] += (l - (l >> 1)) * col[rt];
Tree[rt << 1 | 1] += (l >> 1) * col[rt];
col[rt] = 0;
}
}
void Build(int l,int r,int rt)
{
col[rt] = 0;
if(l == r)
{
scanf("%I64d",&Tree[rt]);
return;
}
int mid = (l+r) >> 1;
Build(lson);
Build(rson);
PushUp(rt);
}
void Upgrade(int ll,int rr,int v,int l,int r,int rt)
{
if(ll <= l && r <= rr)
{
Tree[rt] += (v * (r - l + 1));
col[rt] += v;
return;
}
PushDown(rt,r - l + 1);//
int mid = (l + r) >> 1;
if(ll <= mid) Upgrade(ll,rr,v,lson);
if(rr > mid) Upgrade(ll,rr,v,rson);
PushUp(rt);
}
LL Query(int ll,int rr,int l,int r,int rt)
{
LL sum = 0;
if(ll <= l && r <= rr)
{
sum += Tree[rt];
return sum;
}
PushDown(rt,r - l + 1);
int mid = (l + r) >> 1;
if(ll <= mid) sum += Query(ll,rr,lson);
if(rr > mid) sum += Query(ll,rr,rson);
return sum;
}
int main()
{
int n,a,b,c,q;
char temp[2];
//freopen("FileIn.txt","r",stdin);freopen("FileOut.txt","w",stdout);
scanf("%d%d",&n,&q);
Build(1,n,1);
for(int i = 0; i < q; i++)
{
scanf("%s%d%d",&temp,&a,&b);
if(temp[0] == 'C')
{
scanf("%d",&c);
Upgrade(a,b,c,1,n,1);
}
else
{
printf("%I64d\n",Query(a,b,1,n,1));
}
}
//fclose(stdin);fclose(stdout);
return 0;
}