| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 47665 | Accepted: 14016 | |
| Case Time Limit: 2000MS | ||
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
/**********************
Time: 1579MS Memory: 6724KB Author: zyh Status: Accepted **********************/
#include<stdio.h>
#include<string.h>
#define N 100010
struct IntevalTree{
int L,R;
__int64 value,add;
}tree[N*4];
int k;
int a[N];
__int64 sum;
void build(int root,int l,int r){
tree[root].L = l; tree[root].R = r;
tree[root].add = 0;
if(tree[root].L == tree[root].R){
tree[root].value = a[k++];
return;
}
int mid = (l+r)>>1;
build(root<<1,l,mid);
build(root<<1|1,mid+1,r);
tree[root].value = tree[root<<1].value + tree[root<<1|1].value;
}
void change(int root,int l,int r,long long v){
if( tree[root].L == l && tree[root].R == r){ //最后一次不加了，最后结算时再加上
tree[root].add += v;
return;
}
tree[root].value += v*(r-l+1);
int mid = (tree[root].L + tree[root].R)>>1;
if(r<=mid) change(root<<1,l,r,v);
else if(l>mid) change(root<<1|1,l,r,v);
else {
change(root<<1,l,mid,v);
change(root<<1|1,mid+1,r,v);
}
}
long long query(int p, int left, int right) {
int mid,v;
v=p<<1;
mid=(tree[p].L+tree[p].R)>>1;
if(tree[p].L==left&&tree[p].R==right) {
return (tree[p].value+tree[p].add*(tree[p].R-tree[p].L+1));
}
else{//可能区间有增量，需要计算并下移
tree[v].add+=tree[p].add;
tree[v+1].add+=tree[p].add;//增量下移
tree[p].value+=(tree[p].R-tree[p].L+1)*tree[p].add;
tree[p].add=0;
}
if(right<=mid)
return query(v,left,right);
else if(left>=mid+1)
return query(v+1,left,right);
else
return query(v,left,mid)+query(v+1,mid+1,right);
}
int main()
{
int n,q,i,aa,b,c;
char s[2];
while(scanf("%d%d",&n,&q)!=EOF){
for(i=0;i<n;i++) scanf("%d",&a[i]);
k=0; build(1,1,n);
while(q--){
scanf("%s",&s);
if(s[0]=='C'){
scanf("%d%d%d",&aa,&b,&c);
change(1,aa,b,c);
}
else{
scanf("%d%d",&aa,&b);
printf("%I64d\n",query(1,aa,b));
}
}
}
return 0;
}
/*
10 20
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
Q 1 6
Q 1 9
C 1 4 2
C 2 4 2
Q 1 1
Q 2 2
Q 3 3
Q 4 4
Q 5 5
Q 6 6
Q 7 7
Q 1 3
*/