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
The sums may exceed the range of 32-bit integers.
题目对于初学者来说,有点难,主要考察的是成段更新,用到了懒惰标记,做起来还是很好玩的
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL long long #define M 100005 LL add[M<<2]; LL sum[M<<2]; void PushUp(int rt)//求区间和 { sum[rt]=sum[rt<<1]+sum[rt<<1|1]; } void PushDown(int rt,int m)//向下更新 { 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; } return; } void build(int l,int r,int rt)//建树 { if(l==r) { scanf("%lld",&sum[rt]); return; } int m=(l+r)>>1; build(lson); build(rson); PushUp(rt);//区间求和 } LL query(int x,int y,int l,int r,int rt)//求和操作 { if(x<=l&&r<=y) { return sum[rt]; } PushDown(rt,r-l+1);//这个地方很重要,向下更新 int m=(l+r)>>1; LL ret=0; if(x<=m) ret+=query(x,y,lson); if(m<y) ret+=query(x,y,rson); return ret;//在次WA两次 } void update(int x,int y,int z,int l,int r,int rt) { if(x<=l&&r<=y) { add[rt]+=(LL)z; sum[rt]+=(LL)z*(r-l+1); return; } PushDown(rt,r-l+1);//懒惰标记的体现 int m=(l+r)>>1; if(x<=m) update(x,y,z,lson); if(m<y) update(x,y,z,rson); PushUp(rt);//更新之后的区间和 } int main() { int n,q; char ch[2]; int x,y,z; while(scanf("%d%d",&n,&q)!=EOF) { memset(add,0,sizeof(add)); memset(sum,0,sizeof(sum)); build(1,n,1); for(int i=1;i<=q;i++) { scanf("%s",ch); if(ch[0]=='Q') { scanf("%d%d",&x,&y); printf("%lld\n",query(x,y,1,n,1)); } else { scanf("%d%d%d",&x,&y,&z); update(x,y,z,1,n,1); } } } return 0; }