这道题是这样子的……一个区间求和一个区间加幂
之前刚刚学了降幂大法就是为了写这道题:-D
然后就直接肝……当然最后还要展开多一层的1[EXM?]
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#include <cstdio> #include <algorithm> #include <cstring> #define ll long long #define p(x) ('0'<=x&&x<='9') char cc; int C; template <class T> void read( T &x ) { x=0; cc=getchar(); C=1; while(!p(cc)) { if(cc=='-') C=-1; cc=getchar(); } while(p(cc)) { x=x*10+cc-48; cc=getchar(); } x*=C; } using namespace std; #define N 100010 int f[50],tmp; int n,m,p,c; int Arr[N],time[N]; bool b[N]; int pri[N],tot; int calc_phi(int x) { int ret=x; for(int i=0;pri[i]*pri[i]<=x&&i<tot;i++) if(x%pri[i]==0) { ret-=ret/pri[i]; while(x%pri[i]==0) x/=pri[i]; } if(x>1) ret-=(ret/x); return ret; } int ksm(int x,int top,int mod,bool &pp) { int ret=1; bool ccc=false; for(;top;top>>=1) { if(top&1) pp|=ccc|(1LL*ret*x>=mod),ret=1LL*ret*x%mod; ccc=1LL*x*x>=mod; x=1LL*x*x%mod; } return ret; } int solve(int a,int L) { bool pp=(a>=f[L]); int mmp=a%f[L]; for(;L;L--) mmp=ksm(c,mmp+(pp?f[L]:0),f[L-1],pp); return mmp; } struct node { int l,r,c[2],sum; bool is; }pool[N<<2]; int pcnt; #define Lc pool[x].c[0] #define Rc pool[x].c[1] #define L pool[x].l #define R pool[x].r #define SUM pool[x].sum #define TIME time[L] #define IS pool[x].is inline void PushUP(int x) { SUM = pool[Lc].sum+pool[Rc].sum; SUM>=p?SUM-=p:SUM; IS=pool[Lc].is&pool[Rc].is; } void build(int l,int r,int x) { L=l; R=r; if(l==r) { SUM=Arr[l]%p; IS=(time[l]==tmp); return; } int mid = (l+r) >>1; build(l,mid,Lc=++pcnt); build(mid+1,r,Rc=++pcnt); PushUP(x); } void rebuild(const int &l,const int &r,int x) { if(IS) return; if(L==R) { TIME++; SUM=solve(Arr[L],TIME); IS=(TIME==tmp); return; } int mid=(L+R)>>1; if(mid>=l) rebuild(l,r,Lc); if(mid <r) rebuild(l,r,Rc); PushUP(x); } int query(const int& l,const int& r,int x) { if(l<=L&&R<=r) return pool[x].sum; if(R<l||L>r) return 0; int ret=0; int mid = (L+R)>>1; ret+=query(l,r,Lc); ret+=query(l,r,Rc); return ret>=p?ret-p:ret; } void pre() { for(int i=1;i<=n;i++) read(Arr[i]); int x=p; for(int i=2;i*i<=p;i++) { !b[i]?pri[tot++]=i:b[i]; for(int j=0;j<tot&&pri[j]*i<N;j++) { b[pri[j]*i]=1; if(i%pri[j]==0) break; } } while(x>1) f[tmp++]=x,x=calc_phi(x); f[tmp++]=1; f[tmp]=1; build(1,n,0); } int main() { freopen("BZOJ4869.in","r",stdin); read(n); read(m); read(p); read(c); pre(); static int a,l,r; while(m--) { read(a); read(l); read(r); if(a) printf("%d\n",query(l,r,0)); else(rebuild(l,r,0)); } return 0; } |