POJ 3468
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 54392 | Accepted: 16353 | |
| 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
update:成段增减 query:区间求和
#include<cstdio> #include<cstring> #define lson l, m, rt<<1 #define rson m+1, r, rt<< 1|1 #define LL long long #define maxn 111110 LL add[maxn<<2]; LL sum[maxn<<2]; void PushUp(int rt) { sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void PushDown(int rt,int m) { if(add[rt]) { add[rt<<1] += add[rt]; add[rt<<1|1] += add[rt]; sum[rt<<1] += add[rt]*(m-(m>>1)); sum[rt<<1|1] += add[rt] * (m>>1); add[rt] = 0; } } void build(int l,int r,int rt) { add[rt] =0; if(l == r) { scanf("%lld",&sum[rt]); return ; } int m=(l+r)>>1; build (lson); build(rson); PushUp(rt); } void update(int L,int R,int c,int l,int r,int rt) { if(l >=L && r <=R) { add[rt] += c; sum[rt] += (LL)c*(r-l+1); return ; } PushDown(rt,r-l+1); int m=(l+r)>>1; if(m>=L) update(L,R,c,lson); if(m<R) update(L,R,c,rson); PushUp(rt); } LL 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 m =(l+r)>>1; LL ret =0; if(m>=L) ret += query(L,R,lson); if(m<R) ret += query(L,R,rson); return ret; } int main() { int N,Q; while(~scanf("%d%d",&N,&Q)) { memset(add,0,sizeof(add)); memset(sum,0,sizeof(sum)); build(1,N,1); char op; int a,b; long long c; for(int i=0; i<Q; i++) { getchar(); scanf("%c",&op); if(op=='Q') { scanf("%d %d",&a,&b); printf("%lld\n",query(a,b,1,N,1)); } else if(op=='C') { scanf("%d %d %d",&a,&b,&c); update(a,b,c,1,N,1); } } } return 0; }