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poj------(3468)A Simple Problem with Integers(区间更新)

2024-07-16 11:00:00   230人阅读

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 60745 Accepted: 18522
Case Time Limit: 2000MS

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 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
 
代码：
#include<cstdio>#include<cstring>const int maxn=100005;struct node{ int lef,rig; __int64 sum,cnt; int mid(){   return lef+(rig-lef>>1);  }}; node reg[maxn<<2];void Build(int left ,int right,int pos){  reg[pos]=(node){left,right,0,0};  if((left==right))  {    scanf("%I64d",&reg[pos].sum);    return ;  }  int mid=reg[pos].mid();  Build(left,mid,pos<<1);  Build(mid+1,right,pos<<1|1);  reg[pos].sum=reg[pos<<1].sum+reg[pos<<1|1].sum;}void Update(int left,int right,int pos,int val){    if(reg[pos].lef>=left&&reg[pos].rig<=right)    {      reg[pos].cnt+=val;      reg[pos].sum+=val*(reg[pos].rig-reg[pos].lef+1);      return ;    }    if(reg[pos].cnt)    {      reg[pos<<1].cnt+=reg[pos].cnt;      reg[pos<<1|1].cnt+=reg[pos].cnt;      reg[pos<<1].sum+=reg[pos].cnt*(reg[pos<<1].rig-reg[pos<<1].lef+1);      reg[pos<<1|1].sum+=reg[pos].cnt*(reg[pos<<1|1].rig-reg[pos<<1|1].lef+1);      reg[pos].cnt=0;    }    int mid=reg[pos].mid();    if(left<=mid)        Update(left,right,pos<<1,val);    if(right>mid)        Update(left,right,pos<<1|1,val);  reg[pos].sum=reg[pos<<1].sum+reg[pos<<1|1].sum;}__int64 Query(int left,int right,int pos){    if(left<=reg[pos].lef&&reg[pos].rig<=right)    {      return reg[pos].sum;    }    if(reg[pos].cnt)  //再向下更新一次    {      reg[pos<<1].cnt+=reg[pos].cnt;      reg[pos<<1|1].cnt+=reg[pos].cnt;      reg[pos<<1].sum+=reg[pos].cnt*(reg[pos<<1].rig-reg[pos<<1].lef+1);      reg[pos<<1|1].sum+=reg[pos].cnt*(reg[pos<<1|1].rig-reg[pos<<1|1].lef+1);      reg[pos].cnt=0;    }    int mid=reg[pos].mid();    __int64 res=0;    if(left<=mid)        res+=Query(left,right,pos<<1);    if(mid<right)        res+=Query(left,right,pos<<1|1);   return res;}int main(){    int n,m,a,b,c;    char ss;    while(scanf("%d%d",&n,&m)!=EOF)    {        Build(1,n,1);      while(m--)       {        getchar();        scanf("%c %d%d",&ss,&a,&b);        if(ss==‘Q‘)            printf("%I64d\n",Query(a,b,1));         else{           scanf("%d",&c);           Update(a,b,1,c);         }       }    }  return 0;}

 


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