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 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
The sums may exceed the range of 32-bit integers.
题意：裸线段树 区间更新/查询
题解：线段树每次写都有不同程度的错误.too vegatable
1.延迟标记的add 都要初始化为0
2.延迟的传递是累加而不是赋值
具体看代码。
1 /****************************** 2 code by drizzle 3 blog: www.cnblogs.com/hsd-/ 4 ^ ^ ^ ^ 5 O O 6 ******************************/ 7 //#include<bits/stdc++.h> 8 #include<iostream> 9 #include<cstring> 10 #include<cstdio> 11 #include<map> 12 #include<algorithm> 13 #include<cmath> 14 #define ll long long 15 #define PI acos(-1.0) 16 #define mod 1000000007 17 using namespace std; 18 struct node 19 { 20 ll l,r; 21 ll add; 22 ll sum; 23 } tree[600005]; 24 void buildtree(ll root,ll left,ll right) 25 { 26 tree[root].l=left; 27 tree[root].r=right; 28 tree[root].add=0;//wa点 所有的延迟标记都要初始化 29 if(left==right)//不能放到下面的if中 30 { 31 32 ll exm; 33 scanf("%lld",&exm); 34 tree[root].sum=exm; 35 return ; 36 } 37 ll mid=(left+right)>>1; 38 buildtree(root<<1,left,mid); 39 buildtree(root<<1|1,mid+1,right); 40 tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum; 41 } 42 void pushdown(ll root) 43 { 44 if(tree[root].add==0) return ; 45 tree[root<<1].add+=tree[root].add;//wa点 这里是增加而不是赋值 46 tree[root<<1|1].add+=tree[root].add; 47 tree[root<<1].sum+=(tree[root<<1].r-tree[root<<1].l+1)*tree[root].add; 48 tree[root<<1|1].sum+=(tree[root<<1|1].r-tree[root<<1|1].l+1)*tree[root].add; 49 tree[root].add=0; 50 } 51 void updata(ll root,ll left,ll right,ll c) 52 { 53 if(tree[root].l==left&&tree[root].r==right) 54 { 55 tree[root].add+=c;//wa点 这里是增加而不是赋值 56 tree[root].sum+=(right-left+1)*c; 57 return ; 58 } 59 pushdown(root); 60 ll mid=(tree[root].l+tree[root].r)>>1; 61 if(right<=mid) 62 updata(root<<1,left,right,c); 63 else 64 { 65 if(left>mid) 66 updata(root<<1|1,left,right,c); 67 else 68 { 69 updata(root<<1,left,mid,c); 70 updata(root<<1|1,mid+1,right,c); 71 } 72 } 73 tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum; 74 } 75 ll query(ll root,ll left,ll right) 76 { 77 if(tree[root].l==left&&tree[root].r==right) 78 { 79 return tree[root].sum; 80 } 81 pushdown(root); 82 ll mid=(tree[root].l+tree[root].r)>>1; 83 if(right<=mid) 84 return query(root<<1,left,right); 85 else 86 { 87 if(left>mid) 88 return query(root<<1|1,left,right); 89 else 90 return query(root<<1,left,mid)+query(root<<1|1,mid+1,right); 91 } 92 } 93 ll n,q; 94 char what; 95 ll l1,r1,ad; 96 int main() 97 { 98 while(~scanf("%lld %lld",&n,&q)) 99 { 100 buildtree(1,1,n); 101 getchar(); 102 for(ll i=1; i<=q; i++) 103 { 104 scanf("%c %lld %lld",&what,&l1,&r1); 105 if(what=='Q') 106 printf("%lld ",query(1,l1,r1)); 107 else 108 { 109 scanf("%lld",&ad); 110 updata(1,l1,r1,ad); 111 } 112 getchar(); 113 } 114 } 115 return 0; 116 } 117 /* 118 10 10 119 1 2 3 4 5 6 7 8 9 10 120 C 1 10 1 121 Q 2 3 122 123 10 10 124 1 2 3 4 5 6 7 8 9 10 125 C 1 5 1 126 Q 4 6 127 */