| 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 -
- 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.
- 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.
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: ",s); 49 for(i=0; i<flag; i++) 50 printf("%lld ",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 }