poj3468 A Simple Problem with Integers（线段树）
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 83200 | Accepted: 25771 | |
| 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
题意：一行序列，每次操作把一个区间里的每个数都加上一个数，或者查询一个区间的和。
分析：一道模板题吧，就每次叠加就行了，只是注意数据范围int会超。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define MAXN 100010
struct node
{
int l,r;
ll s,add;//add为每次加的数
}t[MAXN<<2];
int hh[MAXN];
int n,q;
ll ans;
void build(int l, int r, int i)
{
t[i].l = l;
t[i].r = r;
t[i].add = 0;
if(l == r) return ;
int mid = (l+r)>>1;
build(l, mid, i<<1);
build(mid+1, r, i<<1|1);
t[i].s = t[i<<1].s+t[i<<1|1].s;
}
void update(int l, int r, int add, int i)
{
if(t[i].l>r || t[i].r<l) return ;
if(t[i].l>=l && t[i].r<=r)
{
t[i].s += (t[i].r-t[i].l+1)*add;
t[i].add += add;
return ;
}
if(t[i].add)
{
t[i<<1].s += (t[i<<1].r-t[i<<1].l+1)*t[i].add;
t[i<<1].add += t[i].add;
t[i<<1|1].s += (t[i<<1|1].r-t[i<<1|1].l+1)*t[i].add;
t[i<<1|1].add += t[i].add;
t[i].add = 0;
}
update(l, r, add, i<<1);
update(l, r, add, i<<1|1);
t[i].s = t[i<<1].s+t[i<<1|1].s;
}
void query(int l, int r, int i)
{
if(t[i].l>r || t[i].r<l) return ;
if(t[i].l>=l && t[i].r<=r)
{
ans += t[i].s;
return ;
}
if(t[i].add)
{
t[i<<1].s += (t[i<<1].r-t[i<<1].l+1)*t[i].add;
t[i<<1].add += t[i].add;
t[i<<1|1].s += (t[i<<1|1].r-t[i<<1|1].l+1)*t[i].add;
t[i<<1|1].add += t[i].add;
t[i].add = 0;
}
query(l, r, i<<1);
query(l, r, i<<1|1);
t[i].s = t[i<<1].s+t[i<<1|1].s;
}
int main()
{
int a,b,c;
ll k;
char ch;
while(scanf("%d%d",&n,&q)==2)
{
for(int i=1; i<=n; i++)
scanf("%d",&hh[i]);
build(1, n, 1);
for(int i=1; i<=n; i++)
update(i, i, hh[i], 1);
while(q--)
{
getchar();
scanf("%c",&ch);
if(ch == 'C')
{
scanf("%d%d%d",&a,&b,&c);
update(a, b, c, 1);
}
if(ch == 'Q')
{
ans = 0;
scanf("%d%d",&a,&b);
query(a, b, 1);
printf("%lld\n",ans);
}
}
}
return 0;
}