POJ3468 线段树（区间更新，区间求和，延迟标记）
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 97196 | Accepted: 30348 | |
| 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
The sums may exceed the range of 32-bit integers.
题意：
一串数，有两种操作，在区间a,b内每一个数加c,求区间a,b数的和。
线段树模板题。
代码：
1 #include<iostream> 2 #include<string> 3 #include<cstdio> 4 #include<cstring> 5 #include<algorithm> 6 using namespace std; 7 int n,q; 8 long long sum[400005];//数组多乘4倍。 9 long long add[400005]; 10 void pushup(int rt) 11 { 12 sum[rt]=sum[rt<<1]+sum[rt<<1|1]; 13 } 14 void pushdown(int rt,int lne) 15 { 16 if(add[rt]) 17 { 18 add[rt<<1]+=add[rt]; 19 add[rt<<1|1]+=add[rt]; 20 sum[rt<<1]+=add[rt]*(lne-(lne>>1)); 21 sum[rt<<1|1]+=add[rt]*(lne>>1); 22 add[rt]=0; 23 } 24 } 25 void build(int l,int r,int rt) 26 { 27 add[rt]=0; 28 if(l==r) 29 { 30 scanf("%lld",&sum[rt]); 31 return; 32 } 33 int mid=(l+r)>>1; 34 build(l,mid,rt<<1); 35 build(mid+1,r,rt<<1|1); 36 pushup(rt); 37 } 38 void update(int L,int R,int c,int l,int r,int rt) 39 { 40 if(L<=l&&R>=r) 41 { 42 add[rt]+=c; 43 sum[rt]+=(long long)c*(r-l+1); 44 return; 45 } 46 pushdown(rt,r-l+1); 47 int mid=(l+r)>>1; 48 if(L<=mid) 49 update(L,R,c,l,mid,rt<<1); 50 if(R>mid) 51 update(L,R,c,mid+1,r,rt<<1|1); 52 pushup(rt); 53 } 54 long long query(int L,int R,int l,int r,int rt) 55 { 56 if(L<=l&&R>=r) 57 { 58 return sum[rt]; 59 } 60 pushdown(rt,r-l+1); 61 int mid=(l+r)>>1; 62 long long ans=0; 63 if(L<=mid) 64 ans+=query(L,R,l,mid,rt<<1); 65 if(R>mid) 66 ans+=query(L,R,mid+1,r,rt<<1|1); 67 return ans; 68 } 69 int main() 70 { 71 int a,b,c; 72 scanf("%d%d",&n,&q); 73 build(1,n,1); 74 while(q--){ 75 char ch[3]; 76 scanf("%s",ch); //不能用scanf("%c".....),字符后有空格，可以scanf一个字符数组/cin。 77 if(ch[0]=='Q') 78 { 79 scanf("%d%d",&a,&b); 80 printf("%lld\n",query(a,b,1,n,1)); 81 } 82 else if(ch[0]=='C') 83 { 84 scanf("%d%d%d",&a,&b,&c); 85 update(a,b,c,1,n,1); 86 } 87 } 88 return 0; 89 }
