poj 3468 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K
Case Time Limit: 2000MS

这里是题目链接

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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.

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嗯……线段树的成段更新以及区间求和，主要就是在更新的时候并不全部加到每个节点上，而是加到对应区间的父节点上，并做mark标记，而当query到相关子节点时，再将相应的mark加到子节点上，并分别更新父节点和子节点的mark。

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <cmath>#include <string>#include <queue>#include <stack>#define INF 0x3f3f3f3f#define mem(x,y) memset(x,y,sizeof(x))#define SI(x) scanf("%d",&x)#define MAX 100000+10using namespace std;typedef long long ll;int a[MAX];struct node{    int left,right,mid;    ll sum,mark;}b[MAX<<2];void build(int left,int right,int i){    b[i].left=left;    b[i].right=right;    b[i].mark=0;    b[i].mid=(right+left)>>1;    int mid;    if(left==right)    {        b[i].sum=a[left];        return;    }    mid=(right+left)>>1;    build(left,mid,i<<1);    build(mid+1,right,(i<<1)+1);    b[i].sum=b[i<<1].sum+b[(i<<1)+1].sum;}void add(int left,int right, int num,int i){    if(b[i].left==left&&b[i].right==right)    {      //  b[i].sum+=num;      b[i].mark+=num;      return;    }else {        b[i].sum+=(ll)num*(right-left+1);//        int mid ;//        mid= (b[i].right+ b[i].left)>>1;        if(right<=b[i].mid) add(left,right,num,i<<1);        else if(left>b[i].mid) add(left,right,num,(i<<1)+1);        else{            add(left,b[i].mid,num,i<<1);            add(b[i].mid+1,right,num,(i<<1)+1);        }    }}ll query(int left,int right,int i){   // int mid;    if(b[i].left==left&&b[i].right==right) return b[i].sum+b[i].mark*(ll)(right-left+1);    if(b[i].mark!=0){        b[i<<1].mark+=b[i].mark;        b[(i<<1)+1].mark+=b[i].mark;        b[i].sum+=(ll)(b[i].right-b[i].left+1)*b[i].mark;        b[i].mark=0;    } //   mid=(b[i].left+b[i].right)>>1;    if(right<=b[i].mid) return query(left,right,i<<1);    else if(left>b[i].mid) return query(left,right,(i<<1)+1);    else return query(left,b[i].mid,i<<1)+query(b[i].mid+1,right,(i<<1)+1);}int main(){    int n,q,l,r,val;    char str[10];    while(scanf("%d%d",&n,&q)!=EOF)    {        for(int i=1;i<=n;i++)            SI(a[i]);        build(1,n,1);        for(int i=0;i<q;i++)        {            scanf("%s",str);            if(str[0]=='Q'){                scanf("%d%d",&l,&r);                printf("%lld\n",query(l,r,1));            }else{                scanf("%d%d%d",&l,&r,&val);                add(l,r,val,1);            }        }    }    return 0;}