标签：enter bit 两种 ble code key style field namespace
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 116441 | Accepted: 36178 | |
| 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
Source
#include <cstdio> #include <algorithm> using namespace std; const int N=100005; const int maxn = N*3; long long a[maxn],val[maxn],lazy[maxn]; int n,m,ql,qr,A,B,value; void pushdown(int num,int l) { if(lazy[num]) { lazy[num*2] += lazy[num]; lazy[num*2+1] += lazy[num]; val[num*2] += (long long)lazy[num]*(l-(l/2)); val[num*2+1] += (long long)lazy[num]*(l/2); lazy[num] = 0; } } void build(int num,int l,int r) { if(l==r) { val[num] = a[l]; return ; } int mid = (l+r)/2; build(num*2,l,mid); build(num*2+1,mid+1,r); val[num] = val[num*2]+val[num*2+1]; } long long Findsum(int num,int l,int r) { int mid = (l+r)/2; long long ans = 0; if(ql<=l&&qr>=r) { return val[num]; } else { pushdown(num,r-l+1); if(mid>=ql) { ans += Findsum(num*2,l,mid); } if(mid<qr) { ans += Findsum(num*2+1,mid+1,r); } } return ans; } void update(int num,int l,int r) { if(A<=l&&B>=r) { lazy[num] += (long long)value; val[num] += (long long)value*(r-l+1); return ; } pushdown(num,r-l+1); int mid = (l+r)/2; if(A<=mid) update(num*2,l,mid); if(B>mid) update(num*2+1,mid+1,r); val[num] = val[num*2]+val[num*2+1]; } int main() { scanf("%d %d",&n,&m); for(int i=1;i<=n;i++) { scanf("%lld",&a[i]); } build(1,1,n); char c; for(int i=1;i<=m;i++) { getchar(); scanf("%c",&c); if(c==‘Q‘) { scanf("%d %d",&ql,&qr); printf("%lld\n",Findsum(1,1,n)); } else { scanf("%d %d %d",&A,&B,&value); update(1,1,n); } } }
POJ-3468 A Simple Problem with Integers（线段树、段变化+段查询、模板）
标签：enter bit 两种 ble code key style field namespace
原文地址：http://www.cnblogs.com/WWkkk/p/7366754.html
