poj3468 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 31654 Accepted: 8984Case 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

线段树，额，记得longlong就行了。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;typedef struct{    int l,r,lazy;    long long num,tag;}Tree;Tree tree[500005];void Build(int t,int l,int r){    int mid;    tree[t].l=l;    tree[t].r=r;    tree[t].tag=0;    tree[t].lazy=1;    if (l==r)    {        scanf("%lld",&tree[t].num);        return;    }    mid=(l+r)/2;    Build(2*t+1,l,mid);    Build(2*t+2,mid+1,r);    tree[t].num=tree[2*t+1].num+tree[2*t+2].num;}void Add(int t,int x,int y,long long z){    int l,r,mid;    l=tree[t].l;    r=tree[t].r;    if (x==l && y==r)    {        tree[t].lazy=1;        tree[t].tag+=z;        tree[t].num+=z*(r-l+1);        return;    }    mid=(l+r)/2;    if (tree[t].lazy==1)    {        tree[t].lazy=0;        Add(2*t+1,l,mid,tree[t].tag);        Add(2*t+2,mid+1,r,tree[t].tag);        tree[t].tag=0;    }    if (x<=mid) Add(2*t+1,x,min(y,mid),z);    if (y>mid) Add(2*t+2,max(x,mid+1),y,z);    tree[t].num=tree[2*t+1].num+tree[2*t+2].num;}long long Find(int t,int x,int y){    int mid,l,r;    l=tree[t].l;    r=tree[t].r;    if (l==x && y==r)    {        return tree[t].num;    }    mid=(l+r)/2;    if (tree[t].lazy==1)    {        tree[t].lazy=0;        Add(2*t+1,l,mid,tree[t].tag);        Add(2*t+2,mid+1,r,tree[t].tag);        tree[t].tag=0;    }    long long ans=0;    if (x<=mid) ans+=Find(2*t+1,x,min(mid,y));    if (y>mid) ans+=Find(2*t+2,max(mid+1,x),y);    return ans;}int main(){    int i,j,n,m,x,y;    long long z;    char str[20];    while(scanf("%d%d",&n,&m)!=EOF)    {        Build(0,0,n-1);        while(m--)        {            scanf("%s",str);            if (str[0]=='Q')            {                scanf("%d%d",&x,&y);                printf("%lld\n",Find(0,x-1,y-1));            }            else            {                scanf("%d%d%lld",&x,&y,&z);                Add(0,x-1,y-1,z);            }        }    }    return 0;}