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POJ 3468 A Simple Problem with Integers (线段树区域更新)

2024-07-19 07:49:05   214人阅读

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 62431 Accepted: 19141
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
 
线段树的基础区域更新
  1 #include<cstdio>  2 #include<iostream>  3 #include<cstring>  4 #include<stdlib.h>  5 #include<algorithm>  6 #define LL __int64  7 using namespace std;  8 const int MAXN=100000+10;  9 const int INF=0x3f3f3f3f; 10 int b[MAXN]; 11 struct node 12 { 13     int l,r; 14     LL val,col; 15     int mid() 16     { 17         return (l+r)/2; 18     } 19 }a[MAXN*4]; 20  21 void btree(int step,int l,int r) 22 { 23     a[step].l=l; 24     a[step].r=r; 25     a[step].col=0; 26     if(l==r) 27     { 28         a[step].val=b[l]; 29         //scanf("%d",&a[step].val); 这样输入会WA 30         return ; 31     } 32     int mid=a[step].mid(); 33     btree(step*2,l,mid); 34     btree(step*2+1,mid+1,r); 35     a[step].val=a[step*2].val+a[step*2+1].val; 36 } 37  38 void ptree(int step,int L,int R,LL num) 39 { 40     if(L==a[step].l&&a[step].r==R)//找到符合条件的区域就不更新了，记录下要要更新的值 41     { 42         a[step].col+=num; 43         return ; 44     } 45  46     a[step].val+=(R-L+1)*num;//如果不是符合条件的区域，对当前结点更新 47  48     int mid=a[step].mid(); 49     if(R<=mid) 50         ptree(step*2,L,R,num); 51     else if(L>mid) 52         ptree(step*2+1,L,R,num); 53     else 54     { 55         ptree(step*2,L,mid,num); 56         ptree(step*2+1,mid+1,R,num); 57     } 58 } 59  60 LL qtree(int step,int L,int R) 61 { 62     if(L==a[step].l&&a[step].r==R) 63     { 64         return a[step].val+a[step].col*(R-L+1); 65     } 66  67     a[step].val+=(a[step].r-a[step].l+1)*a[step].col;//对当前结点更新 68      69     int mid=a[step].mid();//对子结点更新 70     ptree(step*2,a[step].l,mid,a[step].col); 71     ptree(step*2+1,mid+1,a[step].r,a[step].col); 72     a[step].col=0; 73  74     if(R<=mid) 75         return qtree(step*2,L,R); 76     else if(L>mid) 77         return qtree(step*2+1,L,R); 78     else 79         return qtree(step*2,L,mid)+qtree(step*2+1,mid+1,R); 80 } 81 int main() 82 { 83     //freopen("in.txt","r",stdin); 84     int n,kase; 85     char ch; 86     int star,en,num; 87     while(scanf("%d %d",&n,&kase)!=EOF) 88     { 89         for(int i=1;i<=n;i++) 90             scanf("%d",&b[i]); 91         btree(1,1,n); 92         while(kase--) 93         { 94             cin>>ch; 95             if(ch==‘C‘) 96             { 97                 scanf("%d %d %d",&star,&en,&num);//num输入%I64d会RE 98                 ptree(1,star,en,num); 99             }100             else101             {102                 scanf("%d %d",&star,&en);103                 printf("%I64d\n",qtree(1,star,en));104             }105         }106     }107     return 0;108 }
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POJ 3468 A Simple Problem with Integers (线段树区域更新)


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