A Simple Problem with Integers
Time Limit: 5000MS
Memory Limit: 131072K
Total Submissions: 53169
Accepted: 15897
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.
初学线段树：点这
代码：
#include <cstdio>
using namespace std;
typedef long long ll; #define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn=111111;
ll add[maxn<<2];
ll sum[maxn<<2]; void PushUp(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
} void PushDown(int rt,int m)
{
if(add[rt])
{
add[rt<<1]+=add[rt];
add[rt<<1|1]+=add[rt];
sum[rt<<1]+=(m-(m>>1))*add[rt];
sum[rt<<1|1]+=(m>>1)*add[rt];
add[rt]=0;
}
} void build(int l,int r,int rt)
{
add[rt]=0;
if(l==r)
{
scanf("%lld",&sum[rt]);//
return;
}
int m=(l+r)>>1;
build(lson);
build(rson);
PushUp(rt);
} void update(int L,int R,int c,int l,int r,int rt)
{
if(L<=l&&R>=r)
{
sum[rt]+=(r-l+1)*c;
add[rt]+=c;
return ;
}
PushDown(rt,r-l+1);
int m=(l+r)>>1;
if(L<=m)
update(L,R,c,lson);
if(R>m)
update(L,R,c,rson);
PushUp(rt);
} ll query(int L,int R,int l,int r,int rt)
{
if(L<=l&&R>=r)
return sum[rt];
PushDown(rt,r-l+1);
int m=(l+r)>>1;
ll res=0;
if(L<=m)
res+=query(L,R,lson);
if(R>m)
res+=query(L,R,rson);
return res;
} int main()
{
int N,Q;
scanf("%d%d",&N,&Q);
build(1,N,1);
while(Q--)
{
char s[6];
int a,b;
scanf("%s%d%d",s,&a,&b);
if(s[0]=='Q')
printf("%lld\n",query(a,b,1,N,1));//
else
{
int c;
scanf("%d",&c);
update(a,b,c,1,N,1);
}
}
return 0;
}