POJ3468 A Simple Problem with Integers —— 线段树 区间修改

题目链接：https://vjudge.net/problem/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.

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 2e18;
 const int MAXN = 1e5+;

 LL sum[MAXN*], addv[MAXN*];

 void push_up(int u)
 {
     sum[u] = sum[u*] + sum[u*+];
 }

 void push_down(int u, int l, int r)
 {
     if(addv[u]!=)
     {
         sum[u*] += 1LL*(r-l++)/*addv[u];
         sum[u*+] += 1LL*(r-l+)/*addv[u];
         addv[u*] += addv[u];
         addv[u*+] += addv[u];
         addv[u] = ;
     }
 }

 void add(int u, int l, int r, int x, int y, int val)
 {
     if(x<=l && r<=y)
     {
         addv[u] += val;
         sum[u] += 1LL*(r-l+)*val;
         return;
     }

     push_down(u, l, r);
     int mid = (l+r)/;
     if(x<=mid) add(u*, l, mid, x, y, val);
     if(y>=mid+) add(u*+, mid+, r, x, y, val);
     push_up(u);
 }

 LL query(int u, int l, int r, int x, int y)
 {
     if(x<=l && r<=y)
         return sum[u];

     push_down(u, l, r);
     LL ret = ;
     int mid = (l+r)/;
     if(x<=mid) ret += query(u*, l, mid, x, y);
     if(y>=mid+) ret += query(u*+, mid+, r, x, y);
     return ret;
 }

 int main()
 {
     int n, m;
     while(scanf("%d%d", &n, &m)!=EOF)
     {
         memset(sum, , sizeof(sum));
         memset(addv, , sizeof(addv));
         char op[]; int a, b, c;
         for(int i = ; i<=n; i++)
         {
             scanf("%d", &a);
             add(, , n, i, i, a);
         }

         for(int i = ; i<=m; i++)
         {
             scanf("%s", op);
             if(op[]=='C')
             {
                 scanf("%d%d%d", &a, &b, &c);
                 add(, , n, a, b, c);
             }
             else
             {
                 scanf("%d%d", &a, &b);
                 printf("%lld\n", query(, , n, a, b));
             }
         }
     }
 }