poj 3468 A Simple Problem with Integers(线段树区间更新)

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 122316 Accepted: 37938Case 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

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly--2007.11.25, Yang Yi

题意：有n个数q次操作，每次操作为Q时，是查询l到r的区间的和。操作为C时，是从l到r的每个数都加上一个值。

思路：线段树区间更新。注意懒惰标记也要开long long，不然会WA。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=100000;long long tr[4*maxn];long long add[4*maxn];void pushup(int i){    tr[i]=tr[i<<1]+tr[i<<1|1];}void pushdown(int i,int len){    if(add[i])    {        add[i<<1]+=add[i];        add[i<<1|1]+=add[i];        tr[i<<1]+=add[i]*(len-len/2);        tr[i<<1|1]+=add[i]*(len/2);        add[i]=0;    }}void build(int i,int l,int r){    if(l==r)    {        scanf("%lld",&tr[i]);        return;    }    int mid=(l+r)/2;    build(i*2,l,mid);    build(i*2+1,mid+1,r);    pushup(i);}void update(int i,int l,int r,int x,int y,int c){    if(x<=l&&r<=y)    {        add[i]+=c;        tr[i]=tr[i]+(r-l+1)*c;        return;    }    pushdown(i,r-l+1);    int mid=(l+r)/2;    if(x<=mid) update(2*i,l,mid,x,y,c);    if(y>mid) update(2*i+1,mid+1,r,x,y,c);    pushup(i);}long long query(int i,int l,int r,int x,int y){    long long sum=0;    if(x<=l&&r<=y)    {        sum+=tr[i];        return sum;    }    pushdown(i,r-l+1);    int mid=(l+r)/2;    if(x<=mid) sum+=query(2*i,l,mid,x,y);    if(y>mid) sum+=query(2*i+1,mid+1,r,x,y);    pushup(i);    return sum;}int main(){    int n,q,x,y,c;    char str[10];    while(~scanf("%d%d",&n,&q))    {        build(1,1,n);        memset(add,0,sizeof(add));        for(int i=1;i<=q;i++)        {            scanf("%s",str);            if(str[0]=='Q')            {                scanf("%d%d",&x,&y);                printf("%lld\n",query(1,1,n,x,y));            }            else            {                scanf("%d%d%d",&x,&y,&c);                update(1,1,n,x,y,c);            }        }    }    return 0;}