poj 3468A 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
#include<stdio.h>
#include<string.h>
#define maxn 100005
#define ll long long
char str[10];
ll a[maxn];
struct node
{
ll l,r,sum,ans;
}b[4*maxn];
void build(ll l,ll r,ll i)
{
ll mid;
b[i].l=l;
b[i].r=r;
b[i].ans=0;
if(b[i].l==b[i].r)
{
b[i].sum=a[l];
return;
}
mid=(l+r)/2;
build(l,mid,2*i);
build(mid+1,r,i*2+1);
b[i].sum=b[i*2].sum+b[i*2+1].sum;
}
void pushdown(int i)
{
if(b[i].ans){
b[i*2].ans+=b[i].ans;
b[i*2+1].ans+=b[i].ans;
b[i*2].sum+=b[i].ans*(b[i*2].r-b[i*2].l+1);
b[i*2+1].sum+=b[i].ans*(b[i*2+1].r-b[i*2+1].l+1);
b[i].ans=0;
}
}
void add(ll l,ll r,ll value,ll i)
{
ll mid;
if(b[i].l==l && b[i].r==r)
{
b[i].ans=b[i].ans+value;
b[i].sum+=(b[i].r-b[i].l+1)*value;
return;
}
pushdown(i);
b[i].sum+=(r-l+1)*value; //这一句写了,下面第二句就不用写了,是同一个意思
mid=(b[i].l+b[i].r)/2;
if(l>mid)
add(l,r,value,i*2+1);
else if(r<=mid)
add(l,r,value,i*2);
else
{
add(l,mid,value,i*2);
add(mid+1,r,value,i*2+1);
}
//b[i].sum=b[i*2].sum+b[i*2+1].sum; ---2
}
ll question(ll l,ll r,ll i)
{
ll mid;
if(b[i].l==l && b[i].r==r)
{
return b[i].sum;
}
pushdown(i);
mid=(b[i].l+b[i].r)/2;
if(l>mid)
return question(l,r,i*2+1);
else if(r<=mid)
return question(l,r,i*2);
else if(l<=mid && r>mid)
return question(l,mid,i*2)+question(mid+1,r,i*2+1);
}
int main()
{
ll n,m,c,d,e,i,j;
while(scanf("%lld%lld",&n,&m)!=EOF)
{
for(i=1;i<=n;i++)
scanf("%lld",&a[i]);
build(1,n,1);
while(m--)
{
scanf("%s",str);
if(str[0]=='Q')
{
scanf("%lld%lld",&c,&d);
printf("%lld\n",question(c,d,1));
}
else if(str[0]=='C')
{
scanf("%lld%lld%lld",&c,&d,&e);
add(c,d,e,1);
}
}
}
return 0;
}
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