D - A Simple Problem with Integers
Time Limit: 5000MS
Memory Limit: 131072KB
64bit IO Format: %I64d & %I64u
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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
Sample Output
Hint
The sums may exceed the range of 32-bit integers.
Memory Limit: 131072KB
64bit IO Format: %I64d & %I64u
[Submit] [Go Back] [Status]
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.
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include<iostream>
using namespace std;
#define MAXN 102000
struct node
{
int l;
int r;
__int64 sum;
__int64 add;
}t[4*MAXN];
__int64 rr[MAXN];
void build(int id,int l,int r)
{
int mid = (l+r)/2;
t[id].l = l;
t[id].r = r;
if(r==l)
{
t[id].sum = rr[l];
t[id].add = 0;
}
else
t[id].add = 0;
if(r>l)
{
build(id*2,l,mid);
build(id*2+1,mid+1,r);
t[id].sum = t[id*2].sum + t[id*2+1].sum;
}
}
void change(int id,int start,int end,__int64 num)
{
if(start==t[id].l && end==t[id].r)
{
t[id].add += num;
return ;
}
t[id].sum+= (end-start+1)*num;
int mid = (t[id].l + t[id].r)/2;
if(t[id].add!=0)
{
t[id].sum += (t[id].r-t[id].l+1)*t[id].add;
change(id*2,t[id].l,mid,t[id].add);
change(id*2+1,mid+1,t[id].r,t[id].add);
t[id].add = 0;
}
if(mid>=end)
{
change(id*2,start,end,num);
}
else if(mid+1<=start)
{
change(id*2+1,start,end,num);
}
else
{
change(id*2,start,mid,num);
change(id*2+1,mid+1,end,num);
}
}
__int64 ans;
void find(int id,int start,int end)
{
if (t[id].l==start && t[id].r == end)
{
ans+=t[id].sum+(t[id].r-t[id].l+1)*t[id].add;
return;
}
if (t[id].add!=0)
{
t[id].sum+= (t[id].r-t[id].l+1)*t[id].add;
t[id*2].add+=t[id].add;
t[id*2+1].add+=t[id].add;
t[id].add = 0;
}
int mid = (t[id].l+t[id].r)/2;
if(mid>=end)
find(id*2,start,end);
else if(start>=mid+1)
find(id*2+1,start,end);
else
{
find(id*2,start,mid);
find(id*2+1,mid+1,end);
}
}
int main()
{
int n,m;
int i,j;
char str[10];
int start,end;
__int64 num;
while (scanf("%d%d",&n,&m)!=EOF)
{
for (i=1;i<=n;i++)
{
scanf("%I64d",&rr[i]);
}
build(1,1,n);
for (i=0;i<m;i++)
{
scanf("%s",str);
if (strcmp(str,"C")==0)
{
scanf("%d%d%I64d",&start,&end,&num);
change(1,start,end,num);
}
else
{
scanf("%d%d",&start,&end);
ans = 0;
find(1,start,end);
printf("%I64d
",ans);
}
}
}
return 0;
}