标签：des style blog http io ar color sp for
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
线段树功能:update:成段增减 query:区间求和
1 #include<cstdio> 2 #include<algorithm> 3 4 #define clr(x,y) memset(x,y,sizeof(x)) 5 #define LL long long 6 #define lson l,m,rt<<1 7 #define rson m+1,r,rt<<1|1 8 9 const int maxn=1e5+3511; 10 using namespace std; 11 12 LL sum[maxn<<2],Lazy[maxn<<2]; 13 14 void PushUp(int rt) 15 { 16 sum[rt]=sum[rt<<1]+sum[rt<<1|1]; 17 } 18 19 void PushDown(int rt,int m) 20 { 21 if(Lazy[rt]) { 22 Lazy[rt<<1]+=Lazy[rt]; 23 Lazy[rt<<1|1]+=Lazy[rt]; 24 sum[rt<<1]+=(m-(m>>1))*Lazy[rt]; 25 sum[rt<<1|1]+=(m>>1)*Lazy[rt]; 26 Lazy[rt]=0; 27 } 28 } 29 30 void build(int l,int r,int rt) 31 { 32 int m; 33 Lazy[rt]=0; 34 if(l==r) { 35 scanf("%lld",&sum[rt]); 36 return; 37 } 38 39 m=(l+r)>>1; 40 build(lson); 41 build(rson); 42 PushUp(rt); 43 } 44 45 void Updata(int L,int R,int c,int l,int r,int rt) 46 { 47 int m; 48 if(L<=l && r<=R) { 49 Lazy[rt]+=c; 50 sum[rt]+=(LL)c*(r-l+1); 51 return; 52 } 53 54 PushDown(rt,r-l+1); 55 m=(l+r)>>1; 56 if(L<=m) Updata(L,R,c,lson); 57 if(R>m) Updata(L,R,c,rson); 58 PushUp(rt); 59 60 } 61 62 63 LL query(int L,int R,int l,int r,int rt) 64 { 65 int m; 66 LL ret=0; 67 if(L<=l && r<=R) { 68 return sum[rt]; 69 } 70 71 PushDown(rt,r-l+1); 72 m=(l+r)>>1; 73 if(L<=m) ret+=query(L,R,lson); 74 if(R>m) ret+=query(L,R,rson); 75 76 return ret; 77 } 78 79 80 81 int main() 82 { 83 int Q,n,a,b,c; 84 char st[10]; 85 86 scanf("%d%d",&n,&Q); 87 build(1,n,1); 88 89 while(Q--) { 90 scanf("%s",st); 91 if(st[0]==‘C‘) { 92 scanf("%d%d%d",&a,&b,&c); 93 Updata(a,b,c,1,n,1); 94 } else { 95 scanf("%d%d",&a,&b); 96 printf("%lld\n",query(a,b,1,n,1)); 97 } 98 99 } 100 101 return 0; 102 }
[POJ] 3468 A Simple Problem with Integers [线段树区间更新求和]
标签：des style blog http io ar color sp for
原文地址：http://www.cnblogs.com/sxiszero/p/4125922.html
