Poj 3468 A Simple Problem with Integers【线段树】

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 107324 Accepted: 33474Case 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

题目大意：

线段树区间更新，区间查询。

Ac代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;typedef long long ll;const int maxn=100000+10;ll sum[maxn*4],add[maxn*4];void pushUp(int k){    sum[k]=sum[k*2]+sum[k*2+1];}void pushDown(int k,int l,int r){    if(add[k])    {        int lc=k*2,rc=k*2+1,m=(l+r)/2;        add[lc]+=add[k];        add[rc]+=add[k];        sum[lc]+=add[k]*(m-l+1);        sum[rc]+=add[k]*(r-m);        add[k]=0;    }}void build(int k,int l,int r){    if(l==r)    {        scanf("%lld",&sum[k]);        add[k]=0;        return ;    }    int m=(l+r)/2;    build(k*2,l,m);    build(k*2+1,m+1,r);    pushUp(k);}void update(int a,int b,ll v,int k,int l,int r){    if(a<=l && r<=b)    {        add[k]+=v;        sum[k]+=v*(r-l+1);        return ;    }    pushDown(k,l,r);    int m=(l+r)/2;    if(a<=m)        update(a,b,v,k*2,l,m);    if(b>m)        update(a,b,v,k*2+1,m+1,r);    pushUp(k);}ll ask(int a,int b,int k,int l,int r){    if(a<=l && r<=b)        return sum[k];    pushDown(k,l,r);    int m=(l+r)/2;    ll res=0;    if(a<=m)        res+=ask(a,b,k*2,l,m);    if(b>m)        res+=ask(a,b,k*2+1,m+1,r);    pushUp(k);    return res;}int main(){    int i,j,n,q,a,b;    char op[3];    scanf("%d%d",&n,&q);    build(1,1,n);    while(q--)    {        scanf("%s%d%d",op,&a,&b);        if(op[0]=='C')        {            ll v;            scanf("%lld",&v);            update(a,b,v,1,1,n);        }        else        {            printf("%lld\n",ask(a,b,1,1,n));        }    }    return 0;}