A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 72265 | Accepted: 22299 | |
| 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 4
Sample Output
4 55 9 15
Hint
The sums may exceed the range of 32-bit integers.
Source
POJ Monthly--2007.11.25, Yang Yi
题目意思：
给一个长度为n的数组，有q组操作，操作有两种:Q l, r 即查询l到r的和。 C l, r, val 即把l到r数组元素都加上val
思路：
线段树经典题目，需要用lazy思想，也就是当把l r加上val时，只标记这个区间lazy=val即可 ，当下次再加上一个val时且在l r之间，那么向下传递。
代码：
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <vector> 6 #include <queue> 7 #include <cmath> 8 #include <set> 9 using namespace std; 10 11 #define N 100005 12 #define ll root<<1 13 #define rr root<<1|1 14 #define mid (a[root].l+a[root].r)/2 15 16 17 int max(int x,int y){return x>y?x:y;} 18 int min(int x,int y){return x<y?x:y;} 19 int abs(int x,int y){return x<0?-x:x;} 20 21 int n; 22 23 struct node{ 24 int l, r; 25 __int64 val, sum; 26 }a[N*4]; 27 28 void build(int l,int r,int root){ 29 a[root].l=l; 30 a[root].r=r; 31 a[root].val=0; 32 if(l==r){ 33 scanf("%I64d",&a[root].sum); 34 return; 35 } 36 build(l,mid,ll); 37 build(mid+1,r,rr); 38 a[root].sum=a[ll].sum+a[rr].sum; 39 } 40 41 void down(int root){ 42 if(a[root].val&&a[root].l!=a[root].r) { 43 a[ll].val+=a[root].val; 44 a[rr].val+=a[root].val; 45 a[ll].sum+=(__int64)(a[ll].r-a[ll].l+1)*a[root].val; 46 a[rr].sum+=(__int64)(a[rr].r-a[rr].l+1)*a[root].val; 47 a[root].val=0; 48 } 49 } 50 51 void update(int l,int r,__int64 val,int root){ 52 if(a[root].l==l&&a[root].r==r){ 53 a[root].val+=val; 54 a[root].sum+=(__int64)(a[root].r-a[root].l+1)*val; 55 return; 56 } 57 down(root); 58 if(l>=a[rr].l) update(l,r,val,rr); 59 else if(r<=a[ll].r) update(l,r,val,ll); 60 else { 61 update(l,mid,val,ll); 62 update(mid+1,r,val,rr); 63 } 64 a[root].sum=a[ll].sum+a[rr].sum; 65 } 66 67 __int64 query(int l,int r,int root){ 68 if(a[root].l==l&&a[root].r==r){ 69 return a[root].sum; 70 } 71 down(root); 72 if(r<=a[ll].r) return query(l,r,ll); 73 else if(l>=a[rr].l) return query(l,r,rr); 74 else return query(l,mid,ll)+query(mid+1,r,rr); 75 } 76 77 void out(int root){ 78 if(a[root].l==a[root].r) { 79 printf("%I64d ",a[root].sum); return; 80 } 81 down(root); 82 out(ll); 83 out(rr); 84 } 85 main() 86 { 87 int i, j, k; 88 int q; 89 while(scanf("%d %d",&n,&q)==2){ 90 build(1,n,1); 91 char s[10]; 92 __int64 w; 93 int l, r; 94 while(q--){ 95 scanf("%s",s); 96 if(strcmp(s,"Q")==0){ 97 scanf("%d %d",&l,&r); 98 printf("%I64d ",query(l,r,1)); 99 } 100 else{ 101 scanf("%d %d %I64d",&l,&r,&w); 102 update(l,r,w,1); 103 // out(1); 104 } 105 } 106 } 107 }