poj 3468 A Simple Problem with Integers 线段树区间更新
2017-04-20 14:30
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A Simple Problem with Integers
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
Sample Output
Hint
The sums may exceed the range of 32-bit integers.
给出n个数q次操作
C代表把a到b间的数分别加c
Q要求输出和
A Simple Problem with Integers
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 63565 | Accepted: 19546 | |
Case Time Limit: 2000MS |
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
The sums may exceed the range of 32-bit integers.
给出n个数q次操作
C代表把a到b间的数分别加c
Q要求输出和
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int MAXN=111111; long long sum[MAXN<<2]; long long lazy[MAXN<<2]; void pushup(int rt) { sum[rt]=sum[rt<<1]+sum[rt<<1|1]; } void pushdown(int rt,int m) { if(lazy[rt]) { lazy[rt<<1]+=lazy[rt]; lazy[rt<<1|1]+=lazy[rt]; sum[rt<<1]+=lazy[rt]*(m-(m>>1)); sum[rt<<1|1]+=lazy[rt]*(m>>1); lazy[rt]=0; } } void build(int l,int r,int rt) { lazy[rt]=0; if(l==r) { scanf("%lld",&sum[rt]); return ; } int mid=(l+r)>>1; build(l,mid,rt<<1); build(mid+1,r,rt<<1|1); pushup(rt); } void update(int L,int R,int c,int l,int r,int rt) { if(l>=L&R>=r) { lazy[rt]+=c; sum[rt]+=(long long)c*(r-l+1); return ; } pushdown(rt,r-l+1); int mid=(l+r)>>1; if(L<=mid) update(L,R,c,l,mid,rt<<1); if(R>mid) update(L,R,c,mid+1,r,rt<<1|1); pushup(rt); } long long query(int L,int R,int l,int r,int rt) { if(l>=L&&R>=r) { return sum[rt]; } pushdown(rt,r-l+1); int mid=(l+r)>>1; long long cnt=0; if(L<=mid) cnt=query(L,R,l,mid,rt<<1); if(R>mid) cnt+=query(L,R,mid+1,r,rt<<1|1); return cnt; } int main() { int n,q; char op[2]; int a,b,c; while(scanf("%d %d",&n,&q)!=EOF) { build(1,n,1); //printf("%d\n",sum[1]); while(q--) { scanf("%s",op); if(op[0]=='Q') { scanf("%d %d",&a,&b); printf("%lld\n",query(a,b,1,n,1)); } else { scanf("%d %d %d",&a,&b,&c); update(a,b,c,1,n,1); } } } return 0; }
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