- 27.34%
Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i].
Unfortunately, the longer he learns, the fewer he gets.
That means, if he reads books from ll to rr, he will get a[l] \times L + a[l+1] \times (L-1) + \cdots + a[r-1] \times 2 + a[r]a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r] (LL is the length of [ ll, rr ] that equals to r - l + 1r−l+1).
Now Ryuji has qq questions, you should answer him:
11. If the question type is 11, you should answer how much knowledge he will get after he reads books [ ll, rr ].
22. If the question type is 22, Ryuji will change the ith book's knowledge to a new value.
Input
First line contains two integers nn and qq (nn, q \le 100000q≤100000).
The next line contains n integers represent a[i]( a[i] \le 1e9)a[i](a[i]≤1e9) .
Then in next qq line each line contains three integers aa, bb, cc, if a = 1a=1, it means question type is 11, and bb, ccrepresents [ ll , rr ]. if a = 2a=2 , it means question type is 22 , and bb, cc means Ryuji changes the bth book' knowledge to cc
Output
For each question, output one line with one integer represent the answer.
样例输入
5 3 1 2 3 4 5 1 1 3 2 5 0 1 4 5
样例输出
10 8
题目来源
题意很好理解,我有两种方法来写这道题目。
第一种就是树状数组,倒的梯形面积。
因为下标是从1开始的,所以r+1。横坐标上面的是a数组,下面的是b数组。
代码:
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<bitset> 6 #include<cassert> 7 #include<cctype> 8 #include<cmath> 9 #include<cstdlib> 10 #include<ctime> 11 #include<deque> 12 #include<iomanip> 13 #include<list> 14 #include<map> 15 #include<queue> 16 #include<set> 17 #include<stack> 18 #include<vector> 19 using namespace std; 20 typedef long long ll; 21 22 const double PI=acos(-1.0); 23 const double eps=1e-6; 24 const ll mod=1e9+7; 25 const int inf=0x3f3f3f3f; 26 const int maxn=1e5+10; 27 const int maxm=1e3+10; 28 #define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 29 30 ll a[maxn],b[maxn],n,q; 31 32 int lowbit(int x) 33 { 34 return x&(-x); 35 } 36 37 void add(ll a[],int x,ll val) 38 { 39 for(int i=x;i<=n;i+=lowbit(i)) 40 a[i]+=val; 41 } 42 43 ll query(ll a[],int x) 44 { 45 ll ans=0; 46 for(int i=x;i>0;i-=lowbit(i)) 47 ans+=a[i]; 48 return ans; 49 } 50 51 int main() 52 { 53 cin>>n>>q; 54 for(int i=1;i<=n;i++){ 55 ll val; 56 cin>>val; 57 add(a,i,val);//单纯的保存,类似前缀和 58 add(b,i,i*val);//这样保存,减去的时候正好满足条件,*L,*(L-1)。。。 59 } 60 while(q--){ 61 ll op,l,r; 62 cin>>op>>l>>r; 63 if(op==2){ 64 ll cnt=query(a,l)-query(a,l-1);//单点更新 65 add(a,l,r-cnt); 66 ll ret=query(b,l)-query(b,l-1);//同上 67 add(b,l,l*r-ret); 68 } 69 else{ 70 ll cnt=(r+1)*(query(a,r)-query(a,l-1));//先算出横坐标的和,然后*(r+1)就是一个大矩形的面积 71 ll ret=query(b,r)-query(b,l-1);//倒着的梯形面积,(因为从1开始的,所以是梯形不是三角形) 72 cout<<cnt-ret<<endl; 73 } 74 } 75 }
还有一种线段树的方法,线段树的就是左儿子+右儿子+两者包围的矩形的面积。
为了好理解,画成三角形。
代码睡醒上完课再贴,想睡觉了,头发要紧,哈哈哈哈哈。