链接：https://www.nowcoder.com/acm/contest/77/H
来源：牛客网
Tree Recovery
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 131072K，其他语言262144K
64bit IO Format: %lld
空间限制：C/C++ 131072K，其他语言262144K
64bit IO Format: %lld
题目描述
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.
输入描述:
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.
输出描述:
You need to answer all Q commands in order. One answer in a line.
输入
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
输出
4 55 9 15
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson rt<<1 4 #define rson rt<<1|1 5 #define ll long long 6 const int maxn=1e5+7; 7 ll a[maxn],add[maxn<<2]; 8 struct node 9 { 10 int left,right,mid; 11 ll sum; 12 }c[maxn<<2]; 13 void build(int l,int r,int rt) 14 { 15 add[rt]=0; 16 c[rt].left=l; 17 c[rt].right=r; 18 c[rt].mid=(l+r)>>1; 19 if(l==r) 20 { 21 c[rt].sum=a[l]; 22 return; 23 } 24 build(l,c[rt].mid,lson); 25 build(c[rt].mid+1,r,rson); 26 c[rt].sum=c[lson].sum+c[rson].sum; 27 } 28 void Push_down(int rt) 29 { 30 if(add[rt]!=0) 31 { 32 add[lson]+=add[rt]; 33 add[rson]+=add[rt]; 34 //这里注意，add[lson],add[rson]此时不一定已经更新了， 35 //因此是在原来的基础上加，不是直接赋值 36 c[lson].sum+=(c[lson].right-c[lson].left+1)*add[rt]; 37 c[rson].sum+=(c[rson].right-c[rson].left+1)*add[rt]; 38 add[rt]=0; 39 } 40 } 41 void update(int l,int r,ll C,int rt) 42 { 43 if(l<=c[rt].left&&r>=c[rt].right) 44 { 45 add[rt]+=C; 46 c[rt].sum+=(c[rt].right-c[rt].left+1)*C; 47 return; 48 } 49 Push_down(rt); 50 if(l<=c[rt].mid)update(l,r,C,lson); 51 if(r>c[rt].mid)update(l,r,C,rson); 52 c[rt].sum=c[lson].sum+c[rson].sum; 53 } 54 ll query(int l,int r,int rt) 55 { 56 if(l<=c[rt].left&&r>=c[rt].right) 57 { 58 return c[rt].sum; 59 } 60 Push_down(rt); 61 int m=(c[rt].left+c[rt].right)>>1; 62 ll ans=0; 63 if(l<=m)ans+=query(l,r,lson); 64 if(r>m) ans+=query(l,r,rson); 65 return ans; 66 } 67 int main() 68 { 69 ll n,Q,i,j; 70 scanf("%lld%lld",&n,&Q); 71 for(i=1;i<=n;i++) 72 scanf("%lld",&a[i]); 73 build(1,n,1); 74 while(Q--) 75 { 76 char t[5];int x,y; 77 scanf("%s%d%d",&t,&x,&y); 78 if(t[0]=='C') 79 { 80 ll z;scanf("%lld",&z); 81 update(x,y,z,1); 82 } 83 else 84 { 85 printf("%lld ",query(x,y,1)); 86 } 87 } 88 return 0; 89 }
1 #include<stdio.h> 2 int main() 3 { 4 int n,m,x,y,sum,i,z,a[100002]; 5 char f; 6 scanf("%d%d",&n,&m); 7 for(i=1;i<=n;i++) 8 scanf("%d",&a[i]); 9 getchar(); 10 while(m--) 11 { 12 sum=0; 13 scanf("%c",&f); 14 15 if(f=='C') 16 {scanf("%d%d%d",&x,&y,&z); 17 for(i=x;i<=y;i++) 18 a[i]+=z; 19 } 20 if(f=='Q') 21 { 22 scanf("%d%d",&x,&y); 23 for(i=x;i<=y;i++) 24 sum+=a[i]; 25 printf("%d ",sum); 26 } 27 getchar(); 28 } 29 return 0; 30 }