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POJ 3468 A Simple Problem with Integers(线段树区间求和)
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
区间求和:
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> typedef long long LL; using namespace std; #define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i ) #define REP( i , n ) for ( int i = 0 ; i < n ; ++ i ) #define CLEAR( a , x ) memset ( a , x , sizeof a ) const int maxn=100000; int num[maxn]; LL sum[maxn<<2],add[maxn<<2]; int N,Q; void pushup(int rs) { sum[rs]=sum[rs<<1]+sum[rs<<1|1]; } void pushdown(int rs,int l) { if(add[rs]) { add[rs<<1]+=add[rs]; add[rs<<1|1]+=add[rs]; sum[rs<<1]+=add[rs]*(l-(l>>1)); sum[rs<<1|1]+=add[rs]*(l>>1); add[rs]=0; } } void build(int rs,int l,int r) { if(l==r) { scanf("%I64d",&sum[rs]); return ; } int mid=(l+r)>>1; build(rs<<1,l,mid); build(rs<<1|1,mid+1,r); pushup(rs); } void update(int c,int x,int y,int l,int r,int rs) { if(l>=x&&r<=y) { add[rs]+=c; sum[rs]+=(LL)c*(r-l+1); return ; } pushdown(rs,r-l+1); int mid=(l+r)>>1; if(x<=mid) update(c,x,y,l,mid,rs<<1); if(y>mid) update(c,x,y,mid+1,r,rs<<1|1); pushup(rs); } LL query(int x,int y,int l,int r,int rs) { if(l>=x&&r<=y) return sum[rs]; pushdown(rs,r-l+1); int mid=(l+r)>>1; LL ans=0; if(x<=mid) ans+=query(x,y,l,mid,rs<<1); if(y>mid) ans+=query(x,y,mid+1,r,rs<<1|1); return ans; } int main() { int x,y,z; std::ios::sync_with_stdio(false); while(~scanf("%d%d",&N,&Q)) { CLEAR(sum,0); CLEAR(add,0); build(1,1,N); char str[2]; while(Q--) { scanf("%s",str); if(str[0]=='C') { scanf("%d%d%d",&x,&y,&z); update(z,x,y,1,N,1); } else { scanf("%d%d",&x,&y); printf("%I64d\n",query(x,y,1,N,1)); } } } return 0; }
POJ 3468 A Simple Problem with Integers(线段树区间求和)