专题二树形结构 J - A Simple Problem with Integers

1. 题目
   

   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

   Inputcopy	Outputcopy
   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	4 55 9 15

   Hint
   The sums may exceed the range of 32-bit integers.
2. 思路
   线段树模板题，模板得不能再模板了，甚至oiwiki上面都有现成的
   就是涉及到字母的输入相当麻烦，scanf好像不能先读一个字符再决定下一步操作，最后用的cin
3. 代码
   

   #include<cstdio>
   #include<iostream>
   #include<string>
   #include<cstring>
   #include<algorithm>
   using namespace std;
   
   long long d[1000000],b[1000000],a[1000005];//区间值，懒惰标记，原始值
   long long n,q;
   
   void build(long long s, long long t, long long p) {
   
   	if (s == t) {
   	    d[p] = a[s];
   	    return;
   	}
   
   	long long m = s + ((t - s) >> 1);
   	build(s, m, p * 2), build(m + 1, t, p * 2 + 1);
   	d[p] = d[p * 2] + d[(p * 2) + 1];
   }
   
   void update(long long l, long long r, long long c, long long s, long long t, long long p) {
   
   	if (l <= s && t <= r) {
   		d[p] += (t - s + 1) * c, b[p] += c;
   	    return;
   	}
   
   	long long m = s + ((t - s) >> 1);
   	if (b[p] && s != t) {
   	    d[p * 2] += b[p] * (m - s + 1), d[p * 2 + 1] += b[p] * (t - m);
   		b[p * 2] += b[p], b[p * 2 + 1] += b[p];
   		b[p] = 0;
   	}
   	if (l <= m) update(l, r, c, s, m, p * 2);
   	if (r > m) update(l, r, c, m + 1, t, p * 2 + 1);
   	d[p] = d[p * 2] + d[p * 2 + 1];
   }
   
   long long getsum(long long l, long long r, long long s, long long t, long long p) {
   
   	if (l <= s && t <= r) return d[p];
   
   	long long m = s + ((t - s) >> 1);
   	if (b[p]) {
   	    d[p * 2] += b[p] * (m - s + 1), d[p * 2 + 1] += b[p] * (t - m);
           b[p * 2] += b[p], b[p * 2 + 1] += b[p];
   		b[p] = 0;
   	}
   	long long sum = 0;
   	if (l <= m) sum = getsum(l, r, s, m, p * 2);
   	if (r > m) sum += getsum(l, r, m + 1, t, p * 2 + 1);
   	return sum;
   }
   
   int main()
   {
   	while(scanf("%lld%lld",&n,&q)!=EOF)
   	{
   		memset(d,0,sizeof(d));
   		memset(b,0,sizeof(b));
   		for(int i=1;i<=n;i++){
   			scanf("%lld",&a[i]);
   		}
   		build(1,n,1);
   		char cc;
   		long long x,y,z;
   		while (q--) {
   			cin >> cc >> x >> y;
   		if (cc == 'Q')
   			cout << getsum(x, y, 1, n, 1) <<endl;
   		else
   			cin >> z, update(x, y, z, 1, n, 1);
   		}
   	}
   }