zcmimi's blog
avatar
zc
2020-02-14 14:21:00
  • 本文总阅读量
查看原题

点击跳转

对于第二问:

看到数据范围可以想到矩阵快速幂求斐波那契数列

矩阵乘法具有分配率: AC+BC=(A+B)C

我们可以想到用线段树维护矩阵区间乘

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?__=1,EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
const int N=100011,P=1000000007;
int n,q;
struct mat{int a[2][2];}A={
    1,0,
    1,0,
},B={
    1,1,
    1,0,
},C={
    1,0,
    0,1
},t,s[N<<2],laz[N<<2];
mat operator+(mat x,mat y){
    mat c;
    for(int i=0;i<2;++i)
        for(int j=0;j<2;++j)
            c.a[i][j]=(x.a[i][j]+y.a[i][j])%P;
    return c;
}
mat operator*(mat x,mat y){
    mat c;clr(c.a,0);
    for(int i=0;i<2;++i)
        for(int j=0;j<2;++j)
            for(int k=0;k<2;++k)
                c.a[i][j]=(c.a[i][j]+1ll*x.a[i][k]*y.a[k][j]%P)%P;
    return c;
}
mat pw(int p){
    mat a=C,b=B;
    while(p){
        if(p&1)a=a*b;
        p>>=1;b=b*b;
    }
    return a;
}
#define ls rt<<1
#define rs rt<<1|1
void pd(int rt){
    laz[ls]=laz[ls]*laz[rt];
    laz[rs]=laz[rs]*laz[rt];
    s[ls]=s[ls]*laz[rt];
    s[rs]=s[rs]*laz[rt];
    laz[rt]=C;
}
void build(int l,int r,int rt){
    laz[rt]=C;
    if(l==r){
        int x;in(x);
        s[rt]=A*pw(x-1);
        return;
    }
    int m=(l+r)>>1;
    build(l,m,ls);build(m+1,r,rs);
    s[rt]=s[ls]+s[rs];
}
void upd(int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R){
        s[rt]=s[rt]*t;
        laz[rt]=laz[rt]*t;
        return;
    }
    int m=(l+r)>>1;
    pd(rt);
    if(L<=m)upd(L,R,l,m,ls);
    if(R>m)upd(L,R,m+1,r,rs);
    s[rt]=s[ls]+s[rs];
}
mat ask(int L,int R,int l,int r,int rt){
    if(L<=l&&r<=R)return s[rt];
    int m=(l+r)>>1;
    pd(rt);
    mat ans;clr(ans.a,0);
    if(L<=m)ans=ask(L,R,l,m,ls);
    if(R>m)ans=ans+ask(L,R,m+1,r,rs);
    return ans;
}
signed main(){
    in(n,q);
    build(1,n,1);
    while(q--){
        int u,l,r,v;
        in(u,l,r);
        if(u==1)
            in(v),t=pw(v),
            upd(l,r,1,n,1);
        else out(ask(l,r,1,n,1).a[0][0],ln);
    }
    flush();
}
LG CF718C Sasha-and-Array
comment评论
Search
search