标签：monthly query lan man limit interval search line ati
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 106771 | Accepted: 33308 | |
| 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
Source
#include <cstdio> #define Max 2000000 struct Tree { long long l,r; long long lazy,dis; }tr[Max]; long long n,m; void up(long long k) { tr[k].dis=tr[k<<1].dis+tr[k<<1|1].dis; } void build(long long k,long long l,long long r) { tr[k].l=l;tr[k].r=r; if(l==r) { scanf("%lld",&tr[k].dis); return; } long long mid=(l+r)>>1; build(k<<1,l,mid); build(k<<1|1,mid+1,r); up(k); } void down(long long k) { if(tr[k].l==tr[k].r) return; tr[k<<1].lazy+=tr[k].lazy; tr[k<<1|1].lazy+=tr[k].lazy; tr[k<<1].dis+=tr[k].lazy*(tr[k<<1].r-tr[k<<1].l+1); tr[k<<1|1].dis+=tr[k].lazy*(tr[k<<1|1].r-tr[k<<1|1].l+1); tr[k].lazy=0; } void section_change(long long k,long long l,long long r,long long v) { if(tr[k].l==l&&tr[k].r==r) { tr[k].dis+=v*(r-l+1); tr[k].lazy+=v; return; } if(tr[k].lazy) down(k); long long mid=(tr[k].l+tr[k].r)>>1; if(l>mid) section_change(k<<1|1,l,r,v); else if(r<=mid) section_change(k<<1,l,r,v); else section_change(k<<1,l,mid,v),section_change(k<<1|1,mid+1,r,v); up(k); } long long section_query(long long k,long long l,long long r) { if(tr[k].l==l&&tr[k].r==r) return tr[k].dis; if(tr[k].lazy) down(k); long long mid=(tr[k].l+tr[k].r)>>1; if(l>mid) return section_query(k<<1|1,l,r); else if(r<=mid) return section_query(k<<1,l,r); else return section_query(k<<1,l,mid)+section_query(k<<1|1,mid+1,r); } int main() { scanf("%lld%lld",&n,&m); build(1,1,n); char ch[4]; for(long long x,y,z;m--;) { scanf("%s",ch+1); switch(ch[1]) { case ‘C‘: { scanf("%lld%lld%lld",&x,&y,&z); section_change(1,x,y,z); break; } case ‘Q‘: { scanf("%lld%lld",&x,&y); printf("%lld\n",section_query(1,x,y)); break; } } } return 0; }
POJ 3468 A Simple Problem with Integers
标签：monthly query lan man limit interval search line ati
原文地址：http://www.cnblogs.com/ruojisun/p/6683823.html
