A Simple Problem with Integers
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 139191 | Accepted: 43086 | |
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
The sums may exceed the range of 32-bit integers.
线段树模板题
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
long long int lazy;
long long int data;
long long int l, r;
};
struct node tree[10000000];
long long int Begin[10000000];
void Buildtree( long long int root, long long int l, long long int r )
{
tree[root].l = l;
tree[root].r = r;
tree[root].lazy = 0;
if( l == r )
tree[root].data = Begin[l];
else
{
long long int mid = ( l + r ) / 2;
Buildtree( root*2+1, l, mid );
Buildtree( root*2+2, mid+1, r );
tree[root].data = tree[root*2+1].data + tree[root*2+2].data;
}
}
void Pushdown( long long int root )
{
if( tree[root].lazy != 0 )
{
tree[root*2+1].lazy += tree[root].lazy;
tree[root*2+2].lazy += tree[root].lazy;
tree[root*2+1].data += (tree[root*2+1].r - tree[root*2+1].l + 1)*tree[root].lazy;
tree[root*2+2].data += (tree[root*2+2].r - tree[root*2+2].l + 1)*tree[root].lazy;
tree[root].lazy = 0;
}
}
void Updata ( long long int root, long long int l, long long int r, long long int z )
{
long long int i = tree[root].l, j = tree[root].r;
if( l <= i && j <= r )
{
tree[root].lazy += z;
tree[root].data += ( j - i + 1 ) * z;
return;
}
Pushdown(root);
long long int mid = ( i + j ) / 2;
if( l <= mid )
Updata( root*2+1, l, r, z );
if( r > mid )
Updata( root*2+2, l, r, z );
tree[root].data = tree[root*2+1].data + tree[root*2+2].data;
return;
}
long long int Query ( long int root, long long int l, long long int r )
{
long long int i = tree[root].l, j = tree[root].r;
if( l <= i && r >= j )
return tree[root].data;
Pushdown( root );
long long int mid = ( i + j ) / 2;
long long int sum = 0;
if( l <= mid )
sum += Query( root*2+1, l, r );
if( r > mid )
sum += Query( root*2+2, l, r );
return sum;
}
int main()
{
long long int i, n, q;
scanf("%lld %lld", &n, &q);
for( i=1; i<=n; i++ )
scanf("%lld", &Begin[i]);
Buildtree( 0, 1, n );
while( q-- )
{
char order;
long long int a, b, c;
getchar();
scanf("%c", &order);
if( order == 'C' )
{
scanf("%lld %lld %lld", &a, &b, &c);
Updata( 0, a, b, c);
}
else if( order == 'Q' )
{
scanf("%lld %lld", &a, &b);
printf("%lld\n", Query( 0, a, b ));
}
}
return 0;
}