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
Hint
The sums may exceed the range of 32-bit integers.
Source
POJ Monthly--2007.11.25, Yang Yi
思路：简单的区间更新跟区间求和，注意会爆int
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#define true ture
#define false flase
using namespace std;
#define ll long long
int scan()
{
int res = 0 , ch ;
while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
{
if( ch == EOF ) return 1 << 30 ;
}
res = ch - '0' ;
while( ( ch = getchar() ) >= '0' && ch <= '9' )
res = res * 10 + ( ch - '0' ) ;
return res ;
}
struct is
{
ll l,r;
ll num;
ll lazy;
}tree[200010*3];
void build_tree(ll l,ll r,ll pos)
{
tree[pos].l=l;
tree[pos].r=r;
tree[pos].lazy=0;
if(l==r)
{
//tree[pos].num=1;
scanf("%lld",&tree[pos].num);
return;
}
ll mid=(l+r)/2;
build_tree(l,mid,pos*2);
build_tree(mid+1,r,pos*2+1);
tree[pos].num=tree[pos*2].num+tree[pos*2+1].num;
}
void update(ll l,ll r,ll change,ll pos)
{
if(tree[pos].l==l&&tree[pos].r==r)
{
tree[pos].lazy+=change;
tree[pos].num+=(tree[pos].r-tree[pos].l+1)*change;
return;
}
if(tree[pos].lazy)
{
tree[pos*2].num+=(tree[pos*2].r+1-tree[pos*2].l)*tree[pos].lazy;
tree[pos*2+1].num+=(tree[pos*2+1].r+1-tree[pos*2+1].l)*tree[pos].lazy;
tree[pos*2].lazy+=tree[pos].lazy;
tree[pos*2+1].lazy+=tree[pos].lazy;
tree[pos].lazy=0;
}
ll mid=(tree[pos].l+tree[pos].r)/2;
if(r<=mid)
update(l,r,change,pos*2);
else if(l>mid)
update(l,r,change,pos*2+1);
else
{
update(l,mid,change,pos*2);
update(mid+1,r,change,pos*2+1);
}
tree[pos].num=tree[pos*2].num+tree[pos*2+1].num;
}
ll query(ll l,ll r,ll pos)
{
//cout<<l<<" "<<r<<" "<<pos<<endl;
if(tree[pos].l==l&&tree[pos].r==r)
return tree[pos].num;
if(tree[pos].lazy)
{
tree[pos*2].num+=(tree[pos*2].r+1-tree[pos*2].l)*tree[pos].lazy;
tree[pos*2+1].num+=(tree[pos*2+1].r+1-tree[pos*2+1].l)*tree[pos].lazy;
tree[pos*2].lazy+=tree[pos].lazy;
tree[pos*2+1].lazy+=tree[pos].lazy;
tree[pos].lazy=0;
}
ll mid=(tree[pos].l+tree[pos].r)/2;
if(l>mid)
return query(l,r,pos*2+1);
else if(r<=mid)
return query(l,r,pos*2);
else
return query(l,mid,pos*2)+query(mid+1,r,pos*2+1);
}
int main()
{
ll x,q,i,t;
while(~scanf("%lld",&x))
{
scanf("%lld",&q);
build_tree(1,x,1);
while(q--)
{
char a[10];
ll l,r;
ll change;
scanf("%s%lld%lld",a,&l,&r);
if(a[0]=='C')
{
scanf("%lld",&change);
update(l,r,change,1);
}
else
printf("%lld\n",query(l,r,1));
}
}
return 0;
}
View Code
