POJ 3468 A Simple Problem with Integers（线段树区间修改）

• 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.

  \

  \

  两种操作，

  1.一段连续区间都加上一个数。

  2.查询一段区间的数字之和。

  第一次写区间修改，套错了模板，各种错误层出不穷，关键还是对线段树的理解不够深刻吧。

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  #include<iostream>
  #include<cstdio>
  #define ll long long
  #include<cstring>
  const int MX = 1e5+5;
  ll sum[4 * MX];
  ll la[4 * MX];
  void pushdown(int rt, int l, int r){
  	if(la[rt]){
  		int m = (l + r) / 2;
  		la[2 * rt] +=  la[rt];
  		la[2 * rt + 1] +=  la[rt];						//两端更新 
  		sum[2 * rt] += 1LL * (m - l + 1) * la[rt];			//是+= 
  		sum[2 * rt + 1] += 1LL * (r - m) * la[rt];
  		la[rt] = 0;
  	}
  }
  void build(int l, int r, int rt){
  	if(l == r) {
  	scanf("%I64d",&sum[rt]);
  	return;
  	}
  	int m = (l + r) / 2;
  	build(l, m, 2 * rt);
  	build(m+1, r, 2 * rt +1);
  	sum[rt] = sum[2 * rt] + sum[2 * rt + 1];
  } 
  void update (int st, int en, int co, int l, int r, int rt){
  	if(st <= l && r <= en){
  		sum[rt] += 1LL* (r - l + 1) * co;		//这里错写成 l-r
  		la [rt] += co;							//+= 错写 = 
  		return ;
  	}
  	pushdown(rt, l, r);
  	int m = (l + r) / 2; 
  	if(st <= m) update(st, en, co, l, m, 2 * rt);
  	if(m < en)  update(st, en, co, m+1, r, 2 * rt + 1);
  	sum[rt] = sum[2 * rt] + sum[2 * rt + 1];
  }
  
  ll sea(int st, int en, int l, int r, int rt){
  	if(st <= l && r <= en) return sum[rt];
  	int m = (l + r) /2 ;
  	ll ans = 0;
  	pushdown(rt, l, r);						//pushdown不能省 
  	if(st <= m) ans += sea(st, en, l, m, 2 * rt);
  	if(m < en) ans += sea(st, en, m+1, r, 2 * rt + 1);
  	return ans;
  }
  
  
  int main(){
  	int n,q,a,b,c;
  	char cmd[3];
  	while(scanf("%d%d",&n,&q) != EOF){
  		build(1, n, 1);
  		memset(la, 0, sizeof(la));					//差点sizeof(0);
  		while(q--){
  			scanf("%s",cmd);
  			if(cmd[0] == 'Q'){
  				scanf("%d%d",&a,&b);
  				printf("%I64d\n",sea(a, b, 1, n, 1));
  			}
  			else {
  			scanf("%d%d%d",&a, &b, &c);
  			update(a, b, c, 1, n, 1);
  			}
  		}
  	}
  	return 0;
  } 
  

  
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