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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
Source
#include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> #define true ture #define false flase using namespace std; #define ll long long int scan() { int res = 0 , ch ; while( !( ( ch = getchar() ) >= ‘0‘ && ch <= ‘9‘ ) ) { if( ch == EOF ) return 1 << 30 ; } res = ch - ‘0‘ ; while( ( ch = getchar() ) >= ‘0‘ && ch <= ‘9‘ ) res = res * 10 + ( ch - ‘0‘ ) ; return res ; } struct is { ll l,r; ll num; ll lazy; }tree[200010*3]; void build_tree(ll l,ll r,ll pos) { tree[pos].l=l; tree[pos].r=r; tree[pos].lazy=0; if(l==r) { //tree[pos].num=1; scanf("%lld",&tree[pos].num); return; } ll mid=(l+r)/2; build_tree(l,mid,pos*2); build_tree(mid+1,r,pos*2+1); tree[pos].num=tree[pos*2].num+tree[pos*2+1].num; } void update(ll l,ll r,ll change,ll pos) { if(tree[pos].l==l&&tree[pos].r==r) { tree[pos].lazy+=change; tree[pos].num+=(tree[pos].r-tree[pos].l+1)*change; return; } if(tree[pos].lazy) { tree[pos*2].num+=(tree[pos*2].r+1-tree[pos*2].l)*tree[pos].lazy; tree[pos*2+1].num+=(tree[pos*2+1].r+1-tree[pos*2+1].l)*tree[pos].lazy; tree[pos*2].lazy+=tree[pos].lazy; tree[pos*2+1].lazy+=tree[pos].lazy; tree[pos].lazy=0; } ll mid=(tree[pos].l+tree[pos].r)/2; if(r<=mid) update(l,r,change,pos*2); else if(l>mid) update(l,r,change,pos*2+1); else { update(l,mid,change,pos*2); update(mid+1,r,change,pos*2+1); } tree[pos].num=tree[pos*2].num+tree[pos*2+1].num; } ll query(ll l,ll r,ll pos) { //cout<<l<<" "<<r<<" "<<pos<<endl; if(tree[pos].l==l&&tree[pos].r==r) return tree[pos].num; if(tree[pos].lazy) { tree[pos*2].num+=(tree[pos*2].r+1-tree[pos*2].l)*tree[pos].lazy; tree[pos*2+1].num+=(tree[pos*2+1].r+1-tree[pos*2+1].l)*tree[pos].lazy; tree[pos*2].lazy+=tree[pos].lazy; tree[pos*2+1].lazy+=tree[pos].lazy; tree[pos].lazy=0; } ll mid=(tree[pos].l+tree[pos].r)/2; if(l>mid) return query(l,r,pos*2+1); else if(r<=mid) return query(l,r,pos*2); else return query(l,mid,pos*2)+query(mid+1,r,pos*2+1); } int main() { ll x,q,i,t; while(~scanf("%lld",&x)) { scanf("%lld",&q); build_tree(1,x,1); while(q--) { char a[10]; ll l,r; ll change; scanf("%s%lld%lld",a,&l,&r); if(a[0]==‘C‘) { scanf("%lld",&change); update(l,r,change,1); } else printf("%lld\n",query(l,r,1)); } } return 0; }
poj 3468 A Simple Problem with Integers 线段树加延迟标记
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原文地址：http://www.cnblogs.com/jhz033/p/5404410.html
