| PDF (English) | Statistics | Forum |
| 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.
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 } printf("Case %d:\n",s);
48 for(i=0;i<flag;i++)
49 printf("%lld\n",aa[i]);
50 }
51 }
52 void down(int k)
53 {
54 cnt[2*k+1]+=cnt[k];
55 cnt[2*k+2]+=cnt[k];
56 cnt[k]=0;
57 }
58 void up(int k,int v)
59 {
60 int cc=k;
61 tree[cc]+=cnt[k]*v;
62 if(cc==0)return ;
63 while(cc>=0)
64 {
65 cc=(cc-1)/2;
66 tree[cc]=tree[2*cc+1]+tree[2*cc+2];
67 if(cc==0)
68 return ;
69 }
70 }
71 LL ask(int l,int r,int k,int n,int m)
72 {
73 if(l>m||r<n)
74 { if(cnt[k]>0)
75 {
76 up(k,m-n+1);
77 down(k);
78 }
79 return 0;
80 }
81 else if(l<=n&&r>=m)
82 {
83 if(cnt[k]>0)
84 {
85 up(k,m-n+1);
86 down(k);
87 return tree[k];
88 }
89 else return tree[k];
90 }
91 else
92 {
93 if(cnt[k]>0)
94 {
95 up(k,m-n+1);
96 down(k);
97 }
98 LL nx=ask(l,r,2*k+1,n,(n+m)/2);
99 LL ny=ask(l,r,2*k+2,(n+m)/2+1,m);
100 return nx+ny;
101 }
102 }
103 void in(int l,int r,int k,int n,int m,int u)
104 {
105 if(l>m||r<n)
106 { if(cnt[k]>0)
107 {
108 up(k,m-n+1);
109 down(k);
110 }
111 return ;
112 }
113 else if(l<=n&&r>=m)
114 {
115 cnt[k]+=u;
116 up(k,m-n+1);
117 down(k);
118 return ;
119 }
120 else
121 {
122 if(cnt[k]>0)
123 {
124 up(k,m-n+1);
125 down(k);
126 }
127 in(l,r,2*k+1,n,(n+m)/2,u);
128 in(l,r,2*k+2,(n+m)/2+1,m,u);
129 }
130 }