POJ--3468 A Simple Problem with Integers(线段树)
2015-12-30 22:42
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A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 83559 Accepted: 25862
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
线段树的简单应用。
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 83559 Accepted: 25862
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
线段树的简单应用。
#include <iostream> #include <math.h> #include <string.h> #include <algorithm> #include <stdlib.h> using namespace std; #define MAX 100000 long long int segTree[MAX*4+5]; long long int add[MAX*4+5]; int n,q; char a; int x,y,z; void PushUp(int node) { segTree[node]=segTree[node<<1]+segTree[node<<1|1]; } void PushDown(int node,int m) { if(add[node]!=0) { add[node<<1]+=add[node]; add[node<<1|1]+=add[node]; segTree[node<<1]+=add[node]*(m-(m>>1)); segTree[node<<1|1]+=add[node]*(m>>1); add[node]=0; } } void build(int node,int begin,int end) { add[node]=0; if(begin==end) { scanf("%lld",&segTree[node]); return; } int m=(begin+end)>>1; build(node<<1,begin,m); build(node<<1|1,m+1,end); PushUp(node); } void Update(int node,int begin,int end,int left,int right,int num) { if(left<=begin&&end<=right) { add[node]+=num; segTree[node]+=num*(end-begin+1); return; } PushDown(node,end-begin+1); int m=(begin+end)>>1; if(left<=m) Update(node<<1,begin,m,left,right,num); if(right>m) Update(node<<1|1,m+1,end,left,right,num); PushUp(node); } long long int Query(int node,int begin,int end,int left,int right) { if(left<=begin&&end<=right) return segTree[node]; PushDown(node,end-begin+1); int m=(begin+end)>>1; long long int ret=0; if(left<=m) ret+=Query(node<<1,begin,m,left,right); if(right>m) ret+=Query(node<<1|1,m+1,end,left,right); return ret; } int main() { while(scanf("%d%d",&n,&q)!=EOF) { build(1,1,n); long long int ans; for(int i=0;i<q;i++) { ans=0; getchar(); scanf("%c",&a); if(a=='Q') { scanf("%d%d",&x,&y); ans=Query(1,1,n,x,y); printf("%lld\n",ans); } else { scanf("%d%d%d",&x,&y,&z); Update(1,1,n,x,y,z); } } } return 0; }
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