A Simple Problem with Integers
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
1 #include<cstdio> 2 #include<cstring> 3 using namespace std; 4 5 const int maxn=100050; 6 long long int sum[maxn<<2]; 7 long long int add[maxn<<2]; 8 9 struct node 10 { 11 int l,r; 12 }tree[maxn<<2]; 13 14 void PushUp(int rt) 15 { 16 sum[rt]=sum[rt<<1]+sum[rt<<1|1]; 17 } 18 19 void PushDown(int rt,int m) 20 { 21 if(add[rt]) 22 { 23 add[rt<<1]+=add[rt]; 24 add[rt<<1|1]+=add[rt]; 25 sum[rt<<1]+=add[rt]*(m-(m>>1)); 26 sum[rt<<1|1]+=add[rt]*(m>>1); 27 add[rt]=0; 28 } 29 } 30 31 32 void build(int l,int r,int rt) 33 { 34 tree[rt].l=l; 35 tree[rt].r=r; 36 add[rt]=0; 37 if(l==r) 38 { 39 scanf("%lld",&sum[rt]); 40 return; 41 } 42 int mid=(r+l)>>1; 43 build(l,mid,rt<<1); 44 build(mid+1,r,rt<<1|1); 45 PushUp(rt); 46 } 47 48 void update(long long t,int l,int r,int rt) 49 { 50 if(tree[rt].l==l&&tree[rt].r==r) 51 { 52 add[rt]+=t; 53 sum[rt]+=t*(r-l+1); 54 return; 55 } 56 if(tree[rt].l==tree[rt].r)return; 57 PushDown(rt,tree[rt].r-tree[rt].l+1); 58 int mid=(tree[rt].r+tree[rt].l)>>1; 59 if(r<=mid)update(t,l,r,rt<<1); 60 else if(l>mid)update(t,l,r,rt<<1|1); 61 else 62 { 63 update(t,l,mid,rt<<1); 64 update(t,mid+1,r,rt<<1|1); 65 } 66 PushUp(rt); 67 } 68 69 long long int query(int l,int r,int rt) 70 { 71 if(l==tree[rt].l&&r==tree[rt].r) 72 { 73 return sum[rt]; 74 } 75 PushDown(rt,tree[rt].r-tree[rt].l+1); 76 int mid=(tree[rt].l+tree[rt].r)>>1; 77 long long int ans=0; 78 if(r<=mid)ans+=query(l,r,rt<<1); 79 else if(l>mid)ans+=query(l,r,rt<<1|1); 80 else 81 { 82 ans+=query(l,mid,rt<<1); 83 ans+=query(mid+1,r,rt<<1|1); 84 } 85 return ans; 86 } 87 88 int main() 89 { 90 //freopen("in.txt","r",stdin); 91 char com; 92 int i,t,l,r,val,n,m; 93 while(scanf("%d%d",&m,&n)!=EOF) 94 { 95 memset(add,0,sizeof(add)); 96 build(1,m,1); 97 while(n--) 98 { 99 getchar(); 100 scanf("%c",&com); 101 if(com=='Q') 102 { 103 scanf("%d%d",&l,&r); 104 long long int ans=query(l,r,1); 105 printf("%lld ",ans); 106 } 107 else 108 { 109 scanf("%d%d%d",&l,&r,&val); 110 update(val,l,r,1); 111 } 112 } 113 } 114 return 0; 115 }