标签：other add long 代码 利用 and const 标记 dea
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 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4
Sample Output
4 55 9 15
Hint
1 #include<iostream> 2 #include<stdio.h> 3 using namespace std; 4 5 const int MAXN = 500005; 6 int a[MAXN]; 7 long long int tree[MAXN*4]; 8 long long int lazy[MAXN*4]; 9 int N,Q; 10 string s; 11 int x , y, z; 12 void push_down(int l ,int r ,int rt) 13 { 14 int m = (l+r)/2; 15 16 if(lazy[rt]) 17 { 18 tree[rt*2] += lazy[rt]*(m-l+1); 19 tree[rt*2+1] += lazy[rt]*(r-m); 20 lazy[rt*2] += lazy[rt]; 21 lazy[rt*2+1] += lazy[rt]; 22 lazy[rt] = 0; 23 } 24 } 25 void bulid_tree(int l ,int r ,int rt) 26 { 27 if(l==r) 28 { 29 tree[rt] = a[l]; 30 return ; 31 } 32 int m = (l+r)/2; 33 34 bulid_tree(l,m,rt*2); 35 bulid_tree(m+1,r,rt*2+1); 36 tree[rt] = tree[rt*2]+tree[rt*2+1]; 37 } 38 39 long long int Query(int x ,int y ,int l ,int r ,int rt) 40 { 41 long long sum = 0 ; 42 if(x<=l&&r<=y) 43 { 44 return tree[rt]; 45 } 46 int m = (l+r)/2; 47 push_down(l,r,rt); 48 if(x<=m) 49 { 50 sum += Query(x,y,l,m,rt*2); 51 } 52 if(m<y) 53 { 54 sum += Query(x,y,m+1,r,rt*2+1); 55 } 56 return sum; 57 } 58 void Update(int x ,int y ,int k ,int l ,int r ,int rt) 59 { 60 if(x<=l&&y>=r) 61 { 62 tree[rt] += k*(r-l+1); 63 lazy[rt] += k; 64 return ; 65 } 66 push_down(l,r,rt); 67 int m = (l+r)/2; 68 if(x<=m) 69 { 70 Update(x,y,k,l,m,rt*2); 71 } 72 if(y>m) 73 { 74 Update(x,y,k,m+1,r,rt*2+1); 75 } 76 tree[rt] = tree[rt*2]+tree[rt*2+1]; 77 } 78 int main() 79 { 80 scanf("%d%d",&N,&Q); 81 for(int i = 1 ; i <= N;i++) 82 { 83 scanf("%d",&a[i]); 84 } 85 bulid_tree(1,N,1); 86 while(Q--) 87 { 88 cin>>s; 89 if(s[0]==‘Q‘) 90 { 91 scanf("%d%d",&x,&y); 92 printf("%lld\n",Query(x,y,1,N,1)); 93 94 } 95 else 96 if(s[0]==‘C‘) 97 { 98 scanf("%d%d%d",&x,&y,&z); 99 Update(x,y,z,1,N,1); 100 } 101 } 102 return 0; 103 }
POJ - 3468A Simple Problem with Integers （线段树区间更新，区间查询和）
标签：other add long 代码 利用 and const 标记 dea
原文地址：https://www.cnblogs.com/yewanting/p/10800242.html
