commands bool tdi zoj mean second lib script als

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.

題目傳送門 woc考完奧數的我是不是傻了。。。 各種垃圾錯誤 當初我調序列維護的時候挺順的啊 代碼如下：

#include<cmath>
#include<cstdio>
#include
 
<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;
struct trnode{
    int l,r,lc,rc;
    ll c,add;
    bool lazy;
}tr[210000];int trlen;
void bt(int l,int r)
{
    int now=++trlen;
    tr[now].l=l;tr[now].r=r;tr[now].c=0ll;
    tr[now].lc=tr[now].rc=-1;tr[now].add=0ll,tr[now].lazy=false;
    if(l<r)
    {
        int mid=(l+r)/2;
        tr[now].lc=trlen+1;bt(l,mid);
        tr[now].rc=trlen+1;bt(mid+1,r);
    }
}
void update(int x)
{
    int lc=tr[x].lc,rc=tr[x].rc;
    tr[lc].c+=(tr[lc].r-tr[lc].l+1)*tr[x].add;tr[lc].add+=tr[x].add;
    tr[rc].c+=(tr[rc].r-tr[rc].l+1)*tr[x].add;tr[rc].add+=tr[x].add;
    tr[x].add=0;tr[x].lazy=false;
    tr[lc].lazy=1;tr[rc].lazy=1;
}
void change(int now,int l,int r,ll c)
{
    if(tr[now].l==l&&tr[now].r==r){
        tr[now].c+=(tr[now].r-tr[now].l+1)*c;
        tr[now].add+=c;tr[now].lazy=true;
        return ;
    }    
    int lc=tr[now].lc,rc=tr[now].rc,mid=(tr[now].l+tr[now].r)/2;
    if(tr[now].lazy==true)update(now);
    if(r<=mid)change(lc,l,r,c);
    else if(mid+1<=l)change(rc,l,r,c);
    else
    {
        change(lc,l,mid,c);
        change(rc,mid+1,r,c);
    }
    tr[now].c=tr[lc].c+tr[rc].c;
}
ll findsum(int now,int l,int r)
{
    if(tr[now].l==l&&tr[now].r==r)return tr[now].c;
    if(tr[now].lazy==true)update(now);
    int lc=tr[now].lc,rc=tr[now].rc,mid=(tr[now].l+tr[now].r)/2;
    if(mid+1<=l)return findsum(rc,l,r);
    else if(r<=mid)return findsum(lc,l,r);
    else return findsum(lc,l,mid)+findsum(rc,mid+1,r);
}
int n,m;
int main()
{
    scanf("%d%d",&n,&m);
    trlen=0;bt(1,n);
    for(int i=1;i<=n;i++)
    {
        ll x;
        scanf("%lld",&x);
        change(1,i,i,x);
    }
    char st[6];
    while(m--)
    {
        scanf("%s",st+1);
        if(st[1]==‘Q‘){
            int l,r;
            scanf("%d%d",&l,&r);if(l>r)swap(l,r);
            printf("%lld\n",findsum(1,l,r));
        }
        else
        {
            int x,y;ll c;
            scanf("%d%d%lld",&x,&y,&c);if(x>y)swap(x,y);
            change(1,x,y,c);
        }
    }
    return 0;
}

by_lmy

BZOJ3212: Pku3468 A Simple Problem with Integers