A Simple Problem with Integers
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:区间更新,区间查询
思路:线段树
AC代码(c++ac,g++ TL):
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<utility>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#define maxn 100005
#define INF 0x3f3f3f3f
#define LL long long
#define ULL unsigned long long
#define E 1e-8
#define mod 1000000007
#define P pair<int,int>
#define MID(l,r) (l+(r-l)/2)
#define lson(o) (o<<1)
#define rson(o) (o<<1|1)
using namespace std;
LL a[maxn];
struct node{
int l,r;
LL sum,inc;
}tree[maxn<<2];
void build(int o,int l,int r)
{
tree[o].l = l;
tree[o].r = r;
tree[o].inc = 0;
if(l == r){
tree[o].sum = a[l];
return ;
}
int m = MID(l,r);
int lc = lson(o),rc = rson(o);
build(lc,l,m);
build(rc,m+1,r);
tree[o].sum = tree[lc].sum+tree[rc].sum;
}
void update(int o,int l,int r,LL val)
{
if(tree[o].l == l && tree[o].r == r){
tree[o].inc += val;
return ;
}
tree[o].sum += val*(r-l+1);
int m = MID(tree[o].l,tree[o].r);
int lc = lson(o),rc = rson(o);
if(r<=m) update(lc,l,r,val);
else if(l>m) update(rc,l,r,val);
else{
update(lc,l,m,val);
update(rc,m+1,r,val);
}
}
LL ans_sum;
LL query(int o,int l,int r)
{
if(tree[o].l == l && tree[o].r == r){
return tree[o].sum + tree[o].inc *(r-l+1);
}
tree[o].sum += (tree[o].r-tree[o].l+1)*tree[o].inc;
int m = MID(tree[o].l,tree[o].r);
int lc = lson(o),rc = rson(o);
update(lc,tree[o].l,m,tree[o].inc);
update(rc,m+1,tree[o].r,tree[o].inc); //只更新到下一层
tree[o].inc = 0; //清除标记
if(r<=m) ans_sum = query(lc,l,r);
else if(l>m) ans_sum = query(rc,l,r);
else{
ans_sum = query(lc,l,m)+query(rc,m+1,r);
}
return ans_sum;
}
int main()
{
int n,m,x,y;
LL t;
char c;
while(scanf("%d %d",&n,&m)!=EOF){
for(int i=1;i<=n;++i){
scanf("%lld",&a[i]);
}
build(1,1,n);
while(m--){
cin>>c>>x>>y;
if(c == 'Q'){
ans_sum = 0;
printf("%lld\n",query(1,x,y));
}
else{
cin>>t;
update(1,x,y,t);
}
}
}
return 0;
}