1164 - Horrible Queries

1164 - Horrible Queries

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Time Limit: 2 second(s)

Memory Limit: 64 MB

World is getting more evil and it's getting tougher to get into the Evil League of Evil. Since the legendary Bad Horse has retired, now you have to correctly answer the evil questions of Dr. Horrible, who has a PhD in horribleness (but not in Computer Science). You are given an array of n elements, which are initially all 0. After that you will be given q commands. They are -

1. 0 x y v - you have to add v to all numbers in the range of x to y (inclusive), where x and y are two indexes of the array.
2. 1 x y - output a line containing a single integer which is the sum of all the array elements between x and y (inclusive).

The array is indexed from 0 to n - 1.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case contains two integers n (1 ≤ n ≤ 10⁵) and q (1 ≤ q ≤ 50000). Each of the next q lines contains a task in one of the following form:

0 x y v (0 ≤ x ≤ y < n, 1 ≤ v ≤ 1000)

1 x y (0 ≤ x ≤ y < n)

Output

For each case, print the case number first. Then for each query '1 x y', print the sum of all the array elements between x and y.

Sample Input	Output for Sample Input
2 10 5 0 0 9 10 1 1 6 0 3 7 2 0 4 5 1 1 5 5 20 3 0 10 12 1 1 11 12 1 19 19	Case 1: 60 13 Case 2: 2 0

Note

Dataset is huge. Use faster i/o methods.

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PROBLEM SETTER: IQRAM MAHMUD

SPECIAL THANKS: JANE ALAM JAN (DATASET, SOLUTION)

思路：线段树区间更新

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<queue>
  6 #include<stack>
  7 #include<map>
  8 #include<math.h>
  9 using namespace std;
 10 typedef long long LL;
 11 LL tree[6*100005];
 12 LL cnt[6*100005];
 13 LL aa[100005];
 14 void down(int k);
 15 void up(int k,int v);
 16 LL  ask(int l,int r,int k,int n,int m);
 17 void  in(int l,int r,int k,int n,int m,int u);
 18 int main(void)
 19 {
 20         int i,j,k;
 21         scanf("%d",&k);
 22         int s;
 23         int n,m;
 24         for(s=1; s<=k; s++)
 25         {
 26                 scanf("%d %d",&n,&m);
 27                 memset(tree,0,sizeof(tree));
 28                 memset(cnt,0,sizeof(cnt));
 29                 int flag=0;
 30                 while(m--)
 31                 {
 32                         int x,y;
 33                         int xx,yy,zz;
 34                         scanf("%d",&x);
 35                         if(x==0)
 36                         {
 37                                 scanf("%d %d %d",&xx,&yy,&zz);
 38                                 in(xx,yy,0,0,n-1,zz);
 39                         }
 40                         else
 41                         {
 42                                 scanf("%d %d",&xx,&yy);
 43                                 LL as=ask(xx,yy,0,0,n-1);
 44                                 aa[flag++]=as;
 45                         }
 46
 47                 }
 48                 printf("Case %d:\n",s);
 49                 for(i=0; i<flag; i++)
 50                         printf("%lld\n",aa[i]);
 51         }
 52 }
 53 void down(int k)
 54 {
 55         cnt[2*k+1]+=cnt[k];
 56         cnt[2*k+2]+=cnt[k];
 57         cnt[k]=0;
 58 }
 59 void up(int k,int v)
 60 {
 61         int cc=k;
 62         tree[cc]+=cnt[k]*v;
 63         if(cc==0)return ;
 64         while(cc>=0)
 65         {
 66                 cc=(cc-1)/2;
 67                 tree[cc]=tree[2*cc+1]+tree[2*cc+2];
 68                 if(cc==0)
 69                         return ;
 70         }
 71 }
 72 LL  ask(int l,int r,int k,int n,int m)
 73 {
 74         if(l>m||r<n)
 75         {
 76                 if(cnt[k]>0)
 77                 {
 78                         up(k,m-n+1);
 79                         down(k);
 80                 }
 81                 return 0;
 82         }
 83         else if(l<=n&&r>=m)
 84         {
 85                 if(cnt[k]>0)
 86                 {
 87                         up(k,m-n+1);
 88                         down(k);
 89                         return tree[k];
 90                 }
 91                 else return tree[k];
 92         }
 93         else
 94         {
 95                 if(cnt[k]>0)
 96                 {
 97
 98                         down(k);
 99                 }
100                 LL nx=ask(l,r,2*k+1,n,(n+m)/2);
101                 LL ny=ask(l,r,2*k+2,(n+m)/2+1,m);
102                 return nx+ny;
103         }
104 }
105 void  in(int l,int r,int k,int n,int m,int u)
106 {
107         if(l>m||r<n)
108         {
109                 if(cnt[k]>0)
110                 {
111                         up(k,m-n+1);
112                         down(k);
113                 }
114                 return ;
115         }
116         else if(l<=n&&r>=m)
117         {
118                 cnt[k]+=u;
119                 up(k,m-n+1);
120                 down(k);
121                 return ;
122         }
123         else
124         {
125                 if(cnt[k]>0)
126                 {
127
128                         down(k);
129                 }
130                 in(l,r,2*k+1,n,(n+m)/2,u);
131                 in(l,r,2*k+2,(n+m)/2+1,m,u);
132         }
133 }