POJ 3468 A Simple Problem with Integers(区间更新+区间求和)
2015-08-24 15:23
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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.
ac代码:
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 78622 | Accepted: 24231 | |
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.
ac代码:
#include<stdio.h> struct s { int left; int right; __int64 sum; __int64 lazy; }tree[1000010]; __int64 num[100010]; void pushdown(int len,int i) { if(tree[i].lazy) { tree[i*2].lazy+=tree[i].lazy; tree[i*2+1].lazy+=tree[i].lazy; tree[i*2].sum+=tree[i].lazy*(len-(len/2)); tree[i*2+1].sum+=tree[i].lazy*(len/2); tree[i].lazy=0; } } void build(int l,int r,int i) { tree[i].left=l; tree[i].right=r; tree[i].lazy=0; if(l==r) { tree[i].sum=num[l]; } else { int mid=(l+r)/2; build(l,mid,i*2); build(mid+1,r,i*2+1); tree[i].sum=tree[i*2].sum+tree[i*2+1].sum; } } void update(int l,int r,__int64 c,int i) { int mid=(tree[i].right+tree[i].left)/2; if(tree[i].left>=l&&tree[i].right<=r) { tree[i].lazy+=c; tree[i].sum+=c*(tree[i].right-tree[i].left+1); return; } pushdown(tree[i].right-tree[i].left+1,i); if(l<=mid) { update(l,r,c,i*2); } if(mid<r) { update(l,r,c,i*2+1); } tree[i].sum=tree[i*2].sum+tree[i*2+1].sum; } __int64 query(int l,int r,int i) { int mid=(tree[i].left+tree[i].right)/2; __int64 ans=0; if(tree[i].left>=l&&tree[i].right<=r) { return tree[i].sum; } pushdown(tree[i].right-tree[i].left+1,i); if(l<=mid) { ans+=query(l,r,i*2); } if(r>mid) { ans+=query(l,r,i*2+1); } tree[i].sum=tree[i*2].sum+tree[i*2+1].sum; return ans; } int main() { int i; int a,b,n,m; __int64 d; char ch[10]; while(scanf("%d%d",&n,&m)!=EOF) { for(i=1;i<=n;i++) scanf("%I64d",&num[i]); build(1,n,1); for(i=0;i<m;i++) { scanf("%s",ch); if(ch[0]=='Q') { scanf("%d%d",&a,&b); printf("%I64d\n",query(a,b,1)); } else { scanf("%d%d%I64d",&a,&b,&d); update(a,b,d,1); } } } return 0; }
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