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POJ3468_A Simple Problem with Integers(线段树/成段更新)
解题报告
题意:
略
思路:
线段树成段更新,区间求和。
#include <iostream> #include <cstring> #include <cstdio> #define LL long long #define int_now int l,int r,int root using namespace std; LL sum[500000],lazy[500000]; void push_up(int root,int l,int r) { sum[root]=sum[root*2]+sum[root*2+1] + lazy[root]*(r-l+1); } void update(int root,int l,int r,int ql,int qr,LL v) { if(ql>r||qr<l)return; if(ql<=l&&r<=qr) { lazy[root]+=v; sum[root]+=v*(r-l+1); return ; } int mid=(l+r)/2; update(root*2,l,mid,ql,qr,v); update(root*2+1,mid+1,r,ql,qr,v); push_up(root,l,r); } LL q_sum(int root,int l,int r,int ql,int qr,LL add) { if(ql>r||qr<l)return 0; if(ql<=l&&r<=qr)return sum[root]+add*(r-l+1); int mid=(l+r)/2; return q_sum(root*2,l,mid,ql,qr,add+lazy[root])+q_sum(root*2+1,mid+1,r,ql,qr,add+lazy[root]); } int main() { int n,q,i,j,ql,qr; LL a; scanf("%d%d",&n,&q); for(i=1; i<=n; i++) { scanf("%lld",&a); update(1,1,n,i,i,a); } char str[10]; for(i=1; i<=q; i++) { scanf("%s",str); if(str[0]=='Q') { scanf("%d%d",&ql,&qr); printf("%lld\n",q_sum(1,1,n,ql,qr,0)); } else { scanf("%d%d%lld",&ql,&qr,&a); update(1,1,n,ql,qr,a); } } return 0; }
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 60817 | Accepted: 18545 | |
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