A Simple Problem with Integers~POJ - 3468

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... ,A_b.

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

这题也是线段树的模板题我上一篇写的是单点更新 这一篇为区间更新 。

这题很适合线段树入门。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 using namespace std;
 #define maxn 100005
 long long sum[maxn<<],add[maxn<<];
 void pushup(int rt) {
     sum[rt]=sum[rt<<]+sum[rt<<|];
 }
 void pushdown(int rt ,int l) {
     if (add[rt]) {
         add[rt<<]+=add[rt];
         add[rt<<|]+=add[rt];
         sum[rt<<]+=add[rt]*(l-(l>>));
         sum[rt<<|]+=add[rt]*(l>>);
         add[rt]=;
     }
 }
 void build(int l,int r,int rt) {
     if (l==r) {
         scanf("%lld",&sum[rt]);
         return ;
     }
     int m=(l+r)>>;
     build(l,m,rt<<);
     build(m+,r,rt<<|);
     pushup(rt);
 }
 void updata(int x,int y,int z,int l,int r,int rt) {
     if (x<=l && r<=y) {
         add[rt]+=z;
         sum[rt]+=(long long)z*(r-l+);
         return ;
     }
     pushdown(rt,r-l+);
     int m=(l+r)>>;
     if (x<=m) updata(x,y,z,l,m,rt<<);
     if (y>m)  updata(x,y,z,m+,r,rt<<|);
     pushup(rt);
 }
 long long query(int x,int y,int l,int r,int rt) {
     long long ans=;
     if (x<=l && r<=y ) return sum[rt];
     pushdown(rt,r-l+);
     int m=(l+r)>>;
     if (x<=m) ans+=query(x,y,l,m,rt<<);
     if (y>m ) ans+=query(x,y,m+,r,rt<<|);
     return ans;
 }
 int main() {
     int n,q;
     while(scanf("%d%d",&n,&q)!=EOF) {
         build(,n,);
         char b[];
         int x,y,z;
         while(q--) {
             scanf("%s",b);
             if (b[]=='Q') {
                 scanf("%d%d",&x,&y);
                 printf("%lld\n",query(x,y,,n,));
             } else {
                 scanf("%d%d%d",&x,&y,&z);
                 updata(x,y,z,,n,);
             }
         }
     }
     return ;
 }