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
/********************** 线段树 编程上有难度，高级数据结构。 树状数组能解决的，线段树一定能解决 树状数组编程难度远比线段树低，多做线段树能提高自己的代码能力 ****************************/ #include"iostream" #include"cstdio" #include"cstring" #include"algorithm" using namespace std; struct node { int l,r; long long sum; long long inc; node *pl,*pr; } ; node tree[200001]; int cnt=0; int mid(node *p) { return (p->l+p->r)/2; } void buildtree(node *p,int l,int r) { p->l=l; p->r=r; p->sum=0; p->inc=0; if(l==r) return ; cnt++; p->pl=tree+cnt; cnt++; p->pr=tree+cnt; buildtree(p->pl,l,(l+r)/2); buildtree(p->pr,(l+r)/2+1,r); } void insert(node *p,int i,int v) { if(p->l==i&&p->r==i) { p->sum=v; return ; } p->sum+=v; if(i<=(p->l+p->r)/2) insert(p->pl,i,v); else insert(p->pr,i,v); } void add(node *p,int a,int b,long long c) { if(p->l==a&&p->r==b) { p->inc+=c; return ; } p->sum+=c*(b-a+1); if(b<=(p->l+p->r)/2) add(p->pl,a,b,c); else if(a>(p->l+p->r)/2) add(p->pr,a,b,c); else { add(p->pl,a,(p->l+p->r)/2,c); add(p->pr,(p->l+p->r)/2+1,b,c); } } long long querysum(node *p,int a,int b) { if(p->l==a&&p->r==b) return p->sum+p->inc*(b-a+1); p->sum+=(p->r-p->l+1)*p->inc; add(p->pl,p->l,(p->l+p->r)/2,p->inc); add(p->pr,(p->l+p->r)/2+1,p->r,p->inc); p->inc=0; if(b<=(p->l+p->r)/2) return querysum(p->pl,a,b); else if(a>(p->l+p->r)/2) return querysum(p->pr,a,b); else return querysum(p->pl,a,(p->l+p->r)/2)+querysum(p->pr,(p->l+p->r)/2+1,b); } int main() { int n,q,a,b,c; int i,j,k; char cmd[10]; //cin>>n>>q; scanf("%d%d",&n,&q); cnt=0; buildtree(tree,1,n); for(i=1;i<=n;i++) { //cin>>a; scanf("%d",&a); insert(tree,i,a); } for(i=0;i<q;i++) { //cin>>cmd; scanf("%s",cmd); if(cmd[0]==‘C‘) { //cin>>a>>b>>c; scanf("%d%d%d",&a,&b,&c); add(tree,a,b,c); } else { //cin>>a>>b; scanf("%d%d",&a,&b); cout<<querysum(tree,a,b)<<endl; } } return 0; }
A Simple Problem with Integers,布布扣,bubuko.com
A Simple Problem with Integers
原文地址：http://www.cnblogs.com/767355675hutaishi/p/3861569.html
