BZOJ3212: Pku3468 A Simple Problem with Integers

题目描述：

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.

题解：

线段树裸题。。。

代码如下：

#include<cstdio>
#include<string>
using namespace std;
const int maxn=100005;
int n,q;
struct dyt{
int l,r;
long long num,lazy;
}tree[4*maxn];
inline int read(){
int x=0,flg=1; char ch=getchar();
while (ch<'0'||ch>'9') {if (ch=='-') flg=-1; ch=getchar();}
while (ch>='0'&&ch<='9') x=x*10+ch-48,ch=getchar();
return x*flg;
}
void buildtree(int x,int l,int r){
tree[x].l=l,tree[x].r=r; int mid;
if (l==r) return; else mid=(l+r)>>1;
buildtree(x*2,l,mid); buildtree(x*2+1,mid+1,r);
}
void pushdown(int x){
int num=tree[x].lazy;tree[x].num+=1ll*(tree[x].r-tree[x].l+1)*tree[x].lazy,tree[x].lazy=0;
if (tree[x].l==tree[x].r) return;
tree[x*2].lazy+=num,tree[x*2+1].lazy+=num;
}
void put(int x,int l,int r,int num){
pushdown(x);
if (tree[x].l==l&&tree[x].r==r) {tree[x].lazy=1ll*num; return;}
int mid=(tree[x].l+tree[x].r)>>1;
if (r<=mid) put(x*2,l,r,num);
else if (l>mid) put(x*2+1,l,r,num);
else {put(x*2,l,mid,num); put(x*2+1,mid+1,r,num);}
tree[x].num+=1ll*(r-l+1)*num;
}
long long get(int x,int l,int r){
pushdown(x);
if (tree[x].l==l&&tree[x].r==r) {return tree[x].num;}
int mid=(tree[x].l+tree[x].r)>>1;
if (r<=mid) return get(x*2,l,r);
else if (l>mid) return get(x*2+1,l,r);
else return get(x*2,l,mid)+get(x*2+1,mid+1,r);
}
int main(){
n=read(); q=read();
buildtree(1,1,n);
for (int i=1;i<=n;i++) {int x=read(); put(1,i,i,x);}
for (int i=1;i<=q;i++) {
char ch=getchar();
while (ch!='C'&&ch!='Q') ch=getchar();
int x=read(),y=read();
if (ch=='C') {int z=read(); put(1,x,y,z);}
else printf("%lld\n",get(1,x,y));
}
return 0;
}