A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 141093 | Accepted: 43762 | |
| 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
15Hint
The sums may exceed the range of 32-bit integers.
题意
线段树区间更新求和模板题。
给出n个数和m次操作，Q表示查询该区间内的数的和并输出，C表示把该区间内的所有数都加上一个值。
AC代码
抄的师傅的板子，还不是太理解
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ull unsigned long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
const double E=exp(1);
const int maxn=2e5+10;
const int mod=1e9+7;
using namespace std;
struct wzy
{
ll left,right,len;
ll value;
ll lazy;
}p[maxn<<2];
void push_up(ll o)
{
p[o].value=p[lson].value+p[rson].value;
}
void push_down(ll o)
{
if(p[o].lazy)
{
p[lson].lazy+=p[o].lazy;
p[rson].lazy+=p[o].lazy;
p[lson].value+=p[o].lazy*p[lson].len;
p[rson].value+=p[o].lazy*p[rson].len;
p[o].lazy=0;
}
}
void build(ll o,ll l,ll r)
{
p[o].left=l;p[o].right=r;
p[o].len=r-l+1;
p[o].lazy=0;
if(l==r)
{
ll x;
scanf("%lld",&x);
p[o].value=x;
return ;
}
ll mid=(l+r)>>1;
build(lson,l,mid);
build(rson,mid+1,r);
push_up(o);
}
void update(ll o,ll l,ll r,ll v)
{
if(p[o].left>=l&&p[o].right<=r)
{
p[o].lazy+=v;
p[o].value+=v*p[o].len;
return ;
}
push_down(o);
ll mid=(p[o].left+p[o].right)>>1;
if(r<=mid)
update(lson,l,r,v);
else if(l>mid)
update(rson,l,r,v);
else
{
update(lson,l,mid,v);
update(rson,mid+1,r,v);
}
push_up(o);
}
ll query(ll o,ll l,ll r)
{
if(p[o].left>=l&&p[o].right<=r)
return p[o].value;
ll mid=(p[o].left+p[o].right)>>1;
push_down(o);
if(r<=mid)
return query(lson,l,r);
else if(l>mid)
return query(rson,l,r);
else
return query(lson,l,mid)+query(rson,mid+1,r);
}
int main(int argc, char const *argv[])
{
int n,m;
scanf("%d%d",&n,&m);
build(1,1,n);
char ch[5];
ll a,b,c;
while(m--)
{
scanf("%s",ch);
if(ch[0]=='Q')
{
scanf("%lld%lld",&a,&b);
printf("%lld\n",query(1,a,b));
}
else
{
scanf("%lld%lld%lld",&a,&b,&c);
update(1,a,b,c);
}
}
return 0;
}
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