poj3468（线段树）

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 87678 Accepted: 27225Case Time Limit: 2000MS

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 and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, 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 of A_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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915
线段树即可，，数据改成longlong，也是区间更新，单点查询。
#include <stdio.h>#include <algorithm>#include <iostream>using namespace std;#define LL long long#define lson l,mid,i*2#define rson mid+1,r,i*2+1const LL MAXN=100000;LL num[MAXN];struct tree{    LL l,r;    LL sum;    LL lazy;} tr[MAXN*3];void build(LL l,LL r,LL i){    tr[i].l=l;    tr[i].r=r;    tr[i].lazy=0;    if(l==r)    {        tr[i].sum=num[l];        return;    }    LL mid=(l+r)/2;    build(lson);    build(rson);    tr[i].sum=tr[i*2].sum+tr[i*2+1].sum;}void add(LL l,LL r,LL i,LL c){    if(tr[i].l==l&&tr[i].r==r)    {        tr[i].lazy+=c;                return;    }    tr[i].sum+=c*(r-l+1);    LL mid=(tr[i].l+tr[i].r)/2;    if(r<=mid)        add(l,r,i*2,c);    else if(l>mid)        add(l,r,i*2+1,c);    else    {        add(lson,c);        add(rson,c);    }}LL query(LL l,LL r,LL i){    if(tr[i].l==l&&tr[i].r==r)        return tr[i].sum+(r-l+1)*tr[i].lazy;    tr[i].sum+=(tr[i].r-tr[i].l+1)*tr[i].lazy;    LL mid=(tr[i].l+tr[i].r)/2;    add(tr[i].l,mid,i*2,tr[i].lazy);    add(mid+1,tr[i].r,i*2+1,tr[i].lazy);    tr[i].lazy=0;    if(r<=mid)        return query(l,r,i*2);    else if(l>mid)        return query(l,r,i*2+1);    else        return query(lson)+query(rson);}int main(){    LL n,t,a,b,c;    char ch[2];    scanf("%lld%lld",&n,&t);    for(LL i=1; i<=n; i++)        scanf("%lld",&num[i]);    build(1,n,1);    for(LL i=1; i<=t; i++)    {        scanf("%s",ch);        if(ch[0]=='C')        {            scanf("%lld%lld%lld",&a,&b,&c);            add(a,b,1,c);        }        else        {            scanf("%lld%lld",&a,&b);            printf("%lld\n",query(a,b,1));        }    }    return 0;}

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