ACM HDU 3468 A Simple Problem with Integers（线段树） by kuangbin
题目链接：http://poj.org/problem?id=3468
本文作者：kuangbin
博客地址：http://www.cnblogs.com/kuangbin/
题目：
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 22796 | Accepted: 6106 | |
| 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
Source
/*
POJ 3468 A Simple Problem with Integers
题目意思：
给定Q个数：A1,A2,```,AQ,以及可能多次进行下列两个操作：
1）对某个区间Ai```Aj的数都加n（n可变）
2）对某个区间Ai```Aj求和
*/
#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=100000;
int num[MAXN];
struct Node
{
int l,r;//区间的左右端点
long long nSum;//区间上的和
long long Inc;//区间增量的累加
}segTree[MAXN*3];
void Build(int i,int l,int r)
{
segTree[i].l=l;
segTree[i].r=r;
segTree[i].Inc=0;
if(l==r)
{
segTree[i].nSum=num[l];
return;
}
int mid=(l+r)>>1;
Build(i<<1,l,mid);
Build(i<<1|1,mid+1,r);
segTree[i].nSum=segTree[i<<1].nSum+segTree[i<<1|1].nSum;
}
void Add(int i,int a,int b,long long c)//在结点i的区间（a,b）上增加c
{
if(segTree[i].l==a&&segTree[i].r==b)
{
segTree[i].Inc+=c;
return;
}
segTree[i].nSum+=c*(b-a+1);
int mid=(segTree[i].l+segTree[i].r)>>1;
if(b<=mid) Add(i<<1,a,b,c);
else if(a>mid) Add(i<<1|1,a,b,c);
else
{
Add(i<<1,a,mid,c);
Add(i<<1|1,mid+1,b,c);
}
}
long long Query(int i,int a,int b)//查询a-b的总和
{
if(segTree[i].l==a&&segTree[i].r==b)
{
return segTree[i].nSum+(b-a+1)*segTree[i].Inc;
}
segTree[i].nSum+=(segTree[i].r-segTree[i].l+1)*segTree[i].Inc;
int mid=(segTree[i].l+segTree[i].r)>>1;
Add(i<<1,segTree[i].l,mid,segTree[i].Inc);
Add(i<<1|1,mid+1,segTree[i].r,segTree[i].Inc);
segTree[i].Inc=0;
if(b<=mid) return Query(i<<1,a,b);
else if(a>mid) return Query(i<<1|1,a,b);
else return Query(i<<1,a,mid)+Query(i<<1|1,mid+1,b);
}
int main()
{
int n,q;
int i;
int a,b,c;
char ch;
while(scanf("%d%d",&n,&q)!=EOF)
{
for(i=1;i<=n;i++) scanf("%d",&num[i]);
Build(1,1,n);
for(i=1;i<=q;i++)
{
cin>>ch;
if(ch=='C')
{
scanf("%d%d%d",&a,&b,&c);
Add(1,a,b,c);
}
else
{
scanf("%d%d",&a,&b);
printf("%I64d\n",Query(1,a,b));
}
}
}
return 0;
}
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