poj 3468 A Simple Problem with Integers

Description

You have N integers, A₁,A₂, ... , A_N. 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 andQ. 1 ≤ N,Q ≤ 100000.

The second line contains N numbers, the initial values ofA₁, A₂, ... , A_N. -1000000000 ≤A_i ≤ 1000000000.

Each of the next Q lines represents an operation.

"C a b c" means adding c to each ofA_a, A_a+1, ... , A_b. -10000 ≤c ≤ 10000.

"Q a b" means querying the sum of A_a,A_a+1, ... , A_b.

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.

用long long!!!

题解

用线段树做区间修改与区间求和

代码

#include<cstdio>#include<algorithm>#define N 100010#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1using namespace std;typedef long long ll;ll f[N<<2];ll add[N<<2];int n,q;char s[N];void PushUp(int rt){f[rt]=f[rt<<1]+f[rt<<1|1];}void PushDown(int rt,int m){if(add[rt]){add[rt<<1]+=add[rt];add[rt<<1|1]+=add[rt];f[rt<<1]+=add[rt]*(m-(m>>1));f[rt<<1|1]+=add[rt]*(m>>1);add[rt]=0;}}void build(int l,int r,int rt){add[rt]=0;if(l==r){scanf("%lld",&f[rt]);return ;}int mid=(r+l)>>1;build(lson);build(rson);PushUp(rt);}ll query(int L,int R,int l,int r,int rt){if(l>=L&&r<=R)return f[rt];PushDown(rt,r-l+1);int mid=(r+l)>>1;ll ret=0;if(mid>=L)ret+=query(L,R,lson);if(mid<R)ret+=query(L,R,rson);return ret;}void update(int L,int R,int c,int l,int r,int rt){if(l>=L&&r<=R){add[rt]+=c;f[rt]+=(ll)c*(r-l+1);return ;}PushDown(rt,r-l+1);int mid=(l+r)>>1;if(mid>=L)update(L,R,c,lson);if(mid<R)update(L,R,c,rson);PushUp(rt);}int main(){scanf("%d%d",&n,&q);build(1,n,1);while(q--){int x,y,z;scanf("%s",s);if(s[0]=='Q'){scanf("%d%d",&x,&y);printf("%lld\n",query(x,y,1,n,1));}else{scanf("%d%d%d",&x,&y,&z);update(x,y,z,1,n,1);}}return 0;}