标签：des style blog color os 2014
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
1 /* 2 * Author: Joshua 3 * Created Time: 2014年07月16日 星期三 15时19分46秒 4 * File Name: poj3468.cpp 5 */ 6 #include<cstdio> 7 #include<algorithm> 8 #define maxn 100005 9 #define L(x) (x<<1) 10 #define R(x) (x<<1 |1) 11 typedef long long LL; 12 using namespace std; 13 struct node 14 { 15 int l,r; 16 LL sum,mark; 17 } e[maxn <<2]; 18 int n,m; 19 int a[maxn]; 20 LL ans; 21 void built(int t,int l,int r) 22 { 23 e[t].l=l; 24 e[t].r=r; 25 e[t].mark=0; 26 if (l==r) 27 { 28 e[t].sum=a[e[t].l]; 29 return; 30 } 31 int mid=l+r >>1; 32 built(L(t),l,mid); 33 built(R(t),mid+1,r); 34 e[t].sum=e[L(t)].sum+e[R(t)].sum; 35 } 36 37 void add(int t,int l,int r,int v) 38 { 39 if (l==e[t].l && r==e[t].r) 40 { 41 e[t].mark+=v; 42 return; 43 } 44 e[t].sum+=(r-l+1)*v; 45 int mid=e[t].l+e[t].r >>1; 46 if (r>mid) add(R(t),max(mid+1,l),r,v); 47 if (l<=mid) add(L(t),l,min(mid,r),v); 48 } 49 50 void getSum(int t,int l,int r) 51 { 52 ans+=e[t].mark*(r-l+1); 53 if (e[t].l==l && e[t].r==r) 54 { 55 ans+=e[t].sum; 56 return; 57 } 58 int mid=e[t].l+e[t].r >>1; 59 if (r>mid) getSum(R(t),max(mid+1,l),r); 60 if (l<=mid) getSum(L(t),l,min(mid,r)); 61 } 62 void solve() 63 { 64 for (int i=1;i<=n;++i) 65 scanf("%d",&a[i]); 66 built(1,1,n); 67 char c; 68 int x,y,v; 69 for (int i=1;i<=m;++i) 70 { 71 scanf("%c",&c); 72 scanf("%c%d%d",&c,&x,&y); 73 if (c==‘Q‘) 74 { 75 ans=0; 76 getSum(1,x,y); 77 printf("%lld\n",ans); 78 } 79 else 80 { 81 scanf("%d",&v); 82 add(1,x,y,v); 83 } 84 } 85 } 86 int main() 87 { 88 while (scanf("%d%d",&n,&m)==2) 89 solve(); 90 91 return 0; 92 }
poj3468 A Simple Problem with Integers,布布扣,bubuko.com
poj3468 A Simple Problem with Integers
标签：des style blog color os 2014
原文地址：http://www.cnblogs.com/code-painter/p/3853484.html
