A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 55273 | Accepted: 16628 | |
| 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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4
Sample Output
455915
Hint
The sums may exceed the range of 32-bit integers.
Source
, Yang Yi
解题思路：
某个区间内的数都加上同一个数，成段更新。
代码：
#include#include #include using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int maxn=100005;long long sum[maxn<<2];long long add[maxn<<2];int n,k;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<<1|1]=add[rt];//不能写成这样 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]+=(long long)c*(r-l+1); return; } PushDown(rt,r-l+1); int m=(r+l)>>1; if(L<=m) update(L,R,c,lson); if(R>m) update(L,R,c,rson); PushUp(rt);}long long query(int L,int R,int l,int r,int rt)//注意long long类型{ if(L<=l&&r<=R) { return sum[rt]; } PushDown(rt,r-l+1);//查询的时候要向下更新，懒惰标记 int m=(l+r)>>1; long long ans=0;//注意long long if(L<=m) ans+=query(L,R,lson); if(R>m) ans+=query(L,R,rson); return ans;}int main(){ scanf("%d%d",&n,&k); build(1,n,1); char c;int a,b,d; while(k--) { scanf("%s",&c); if(c=='Q') { scanf("%d%d",&a,&b); printf("%lld\n",query(a,b,1,n,1)); } else { scanf("%d%d%d",&a,&b,&d); update(a,b,d,1,n,1); } } return 0;}