A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 75541 Accepted: 23286 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 4Sample Output
4 55 9 15Hint
The sums may exceed the range of 32-bit integers.Source
POJ Monthly--2007.11.25, Yang Yi
朴素的向下更新会产生很多不必要的操作，所以lazy思想便是，用到再更新，不用就暂存到当前区间。
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 #include <math.h> 5 #include <iostream> 6 #include <algorithm> 7 #include <climits> 8 #include <queue> 9 #define ll long long 10 11 using namespace std; 12 13 //lazy思想的线段树区间更新，区间查询，感觉不难 14 15 const int MAX = 200005; 16 ll num[MAX]; 17 18 struct nodes 19 { 20 int left,right; 21 ll large,add; 22 } tree[MAX*5]; 23 24 void pushup(int root)//代替递归向上更新 25 { 26 tree[root].large = tree[root*2].large + tree[root*2+1].large; 27 } 28 void pushdown(int root)//递归向下更新 29 { 30 if(tree[root].add) 31 { 32 tree[root*2].add += tree[root].add; 33 tree[root*2+1].add += tree[root].add; 34 tree[root*2].large += tree[root].add * (tree[root*2].right - tree[root*2].left + 1); 35 tree[root*2+1].large += tree[root].add * (tree[root*2+1].right - tree[root*2+1].left + 1); 36 tree[root].add = 0; 37 } 38 } 39 void build(int root,int left,int right)//建树过程，也是初始化更新过程 40 { 41 tree[root].left = left; 42 tree[root].right = right; 43 tree[root].add = 0;//不能忘记这个值的初始化为0 44 if(left == right) 45 { 46 tree[root].large = num[left]; 47 return;//注意这里的递归结束条件 48 } 49 50 int mid = (left+right)/2; 51 52 build(2*root,left,mid); 53 build(2*root+1,mid+1,right); 54 55 pushup(root);//该子区间建立好后，该区间更新 56 } 57 58 void update(int root,int left,int right,ll val) 59 { 60 if(left == tree[root].left && right == tree[root].right) // 刚好进入到所需区间，就可以更新到当前区间直接返回 61 { 62 tree[root].add += val; // 用当前节点的add变量记录区间每个元素的累加量 63 tree[root].large += val * (right - left + 1);//更新当前区间和 64 return; 65 } 66 pushdown(root); //否则向下更新一层，继续找合适区间 67 int mid = (tree[root].left+tree[root].right)/2; 68 if(left <= mid && right > mid) 69 { 70 update(2*root,left,mid,val); 71 update(2*root+1,mid+1,right,val); 72 } 73 else if(left > mid) 74 { 75 update(2*root+1,left,right,val); 76 } 77 else 78 { 79 update(2*root,left,right,val); 80 } 81 pushup(root); 82 } 83 84 85 ll query(int root ,int left,int right) 86 { 87 if(left == tree[root].left && right == tree[root].right) //如果是合适区间，直接返回节点值 88 { 89 return tree[root].large; 90 } 91 pushdown(root); //否则向下更新一层，继续查找合适区间 92 int mid = (tree[root].left+tree[root].right)/2; 93 ll ans = 0; 94 if(right > mid && left <= mid) 95 { 96 ans += query(2*root,left,mid); 97 ans += query(2*root+1,mid+1,right); 98 } 99 else if(left > mid) 100 { 101 ans += query(2*root+1,left,right); 102 } 103 else 104 { 105 ans += query(2*root,left,right); 106 } 107 return ans; 108 } 109 110 int main(void) 111 { 112 int n,q,i,x,y; 113 ll c; 114 char cmd; 115 scanf("%d %d",&n,&q); 116 for(i = 1; i <= n; i++) 117 scanf("%lld",&num[i]); 118 build(1,1,n); 119 for(i = 0; i < q; i++) 120 { 121 getchar(); 122 scanf("%c",&cmd); 123 if(cmd == 'Q') 124 { 125 scanf("%d %d",&x,&y); 126 printf("%lld ",query(1,x,y)); 127 } 128 else 129 { 130 scanf("%d %d %lld",&x,&y,&c); 131 update(1,x,y,c); 132 } 133 } 134 return 0; 135 }