E3. Summer Homework
time limit per test
memory limit per test
input
output
By the age of three Smart Beaver mastered all arithmetic operations and got this summer homework from the amazed teacher:
a1, a2, ..., an. Your task is to perform on it m
- xiandviassign valuevito elementaxi.
- liandriyou've got to calculate sum
- , wheref0=f1= 1 and ati≥ 2:fi=fi - 1+fi - 2.
- liridiyou should increase valueaxbydifor allx(li≤x≤ri).
Smart Beaver planned a tour around great Canadian lakes, so he asked you to help him solve the given problem.
Input
n and m (1 ≤ n, m ≤ 2·105) — the number of integers in the sequence and the number of operations, correspondingly. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 105). Then follow m lines, each describes an operation. Each line starts with an integer ti (1 ≤ ti ≤ 3) — the operation type:
- ti= 1, then next follow two integersxivi(1 ≤xi≤n, 0 ≤vi≤ 105);
- ti= 2, then next follow two integersliri(1 ≤li≤ri≤n);
- ti= 3, then next follow three integersliridi(1 ≤li≤ri≤n, 0 ≤di≤ 105).
30
- ndoes not exceed 100,mdoes not exceed 10000 and there will be no queries of the 3-rd type.
70
- 1-st and 2-nd type only.
100
- No extra limitations.
Output
1000000000 (109).
Sample test(s)
input
5 5 1 3 1 2 4 2 1 4 2 1 5 2 2 4 1 3 10 2 1 5
output
12 32 8 50
input
5 4 1 3 1 2 4 3 1 4 1 2 2 4 1 2 10 2 1 5
output
12 45
艰难滴完成了E题……感觉不会再爱了
好吧，题目那个解法由于智商太低没看懂……
说一下我最后乱推的结果
设S(0)=F1a1+F2a2+..+Fnan
+)S(1)=F2a2+F3a3+..+F3a3
= (F1+F2)a2+(F2+F3)a3+..+(Fn-1+Fn)an
= F3 a3+ F4 a3+..+ Fn+1 an
于是我XXXXX……
只要存S0和S1就行了 Sx可以用关于S0和S1的表达式表示：Sx=Fn-2*S0 +Fn-1*S1 (系数为什么是Fib数？因为Sx=Sx-1+Sx-2的不明公式,可用数学归纳法证)
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MAXN (200000+10)
#define F (1000000000)
#define Lson (x<<1)
#define Rson ((x<<1)+1)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
//void inv(long long a,long long b){return exgcd(a,}
int n,m;
long long f[MAXN],sf[MAXN]={0},f2[MAXN],sf2[MAXN];
struct arr_t
{
long long a[MAXN*4],s0[MAXN*4],s1[MAXN*4],b[MAXN*4];
arr_t(){MEM(a)MEM(s0)MEM(s1)MEM(b)}
long long S(int x,int n) //求的是S(x)
{
if (n==1) return s0[x];
if (n==2) return s1[x];
return add(mul(f[n-2],s0[x]),mul(f[n-1],s1[x]));
}
void update(int x,int l,int r)
{
int n=r-l+1,m=(l+r)>>1,ls=m-l+1,rs=n-ls;
s0[x]=(s0[Lson]+S(Rson,ls+1))%F;
s1[x]=(s1[Lson]+S(Rson,ls+2))%F;
}
void change(int x,int n,int c)
{
s0[x]=add(s0[x],mul(sf[n],c));
s1[x]=add(s1[x],mul(sf2[n],c));
}
void pushdown(int x,int l,int r)
{
if (l==r) return;
else if (b[x])
{
int n=r-l+1,m=(l+r)>>1,ls=m-l+1,rs=n-ls;
s0[Lson]=add(s0[Lson],mul(sf[ls],b[x]));
s0[Rson]=add(s0[Rson],mul(sf[rs],b[x]));
s1[Lson]=add(s1[Lson],mul(sf2[ls],b[x]));
s1[Rson]=add(s1[Rson],mul(sf2[rs],b[x]));
b[Lson]+=b[x];b[Rson]+=b[x];b[x]=0;
}
}
void ins(int x,int l,int r,int L,int R,int c)
{
pushdown(x,l,r);
int n=r-l+1,m=(l+r)>>1;
if (L<=l&&r<=R) {b[x]+=c;change(x,n,c);return;}
if (L<=m) ins(Lson,l,m,L,R,c);
if (m<R) ins(Rson,m+1,r,L,R,c);
update(x,l,r);
}
void modify(int x,int l,int r,int L,int c)
{
pushdown(x,l,r);
int n=r-l+1,m=(l+r)>>1;
if (l==r) {s0[x]=s1[x]=c;return;}
if (L<=m) modify(Lson,l,m,L,c);
else modify(Rson,m+1,r,L,c);
update(x,l,r);
}
void build(int x,int l,int r)
{
if (l==r) {scanf("%d",&a[x]);s0[x]=s1[x]=a[x]; return ;}
int n=r-l+1,m=(l+r)>>1;
build(Lson,l,m);
build(Rson,m+1,r);
update(x,l,r);
}
long long qur(int x,int l,int r,int L,int R)
{
pushdown(x,l,r);
int n=r-l+1,m=(l+r)>>1;
if (L<=l&&r<=R) {return S(x,l-(L-1));}
long long ans=0;
if (L<=m) ans=add(ans,qur(Lson,l,m,L,R));
if (m<R) ans=add(ans,qur(Rson,m+1,r,L,R));
return ans;
}
}T;
int main()
{
//freopen("homework.in","r",stdin);
//freopen(".out","w",stdout);
scanf("%d%d",&n,&m);
f[1]=f[2]=sf[1]=1,sf[2]=2;Fork(i,3,n) f[i]=(f[i-1]+f[i-2])%F,sf[i]=(sf[i-1]+f[i])%F;
f2[1]=sf2[1]=1,f2[2]=2;sf2[2]=3;Fork(i,3,n) f2[i]=(f2[i-1]+f2[i-2])%F,sf2[i]=(sf2[i-1]+f2[i])%F;
T.build(1,1,n);
For(i,m)
{
int p,x,v,l,r,d;scanf("%d",&p);
if (p==1)
{
scanf("%d%d",&x,&v);T.modify(1,1,n,x,v);
}
else if (p==2)
{
scanf("%d%d",&l,&r);cout<<T.qur(1,1,n,l,r)<<endl;
}
else
{
scanf("%d%d%d",&l,&r,&d);T.ins(1,1,n,l,r,d);
}
}
return 0;
}
