4542: [Hnoi2016] Large number | MO Team

HN one day Test two Mo team is what ghost. Or the team is not the right posture.
Think about that already. L. . R New additions to the answers R+1 How to update the current answer
The suffix needs to be preprocessed first modp The value < Span class= "Mrow" id= "mathjax-span-4232" > b i Span style= "Display:inline-block; width:0px; Height:2.102em; " > , assume that the sub-sequence L. . R Mode p The value is x
So x?10 r l +b[R]=b[L] And then you can just count the team.
When the modulus is 2 or 5, a special sentence is needed.

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <vector>#include <cmath>#include <queue>#include <ctime>#include <set>#include <map>#define N 110000#define LL Long LongUsing namespace Std;struct w{intL,r,p;}Q[n]; chars[N];long long Sum[n],mod,ss[n],ans[n];long long a[n],b[n],c[n],h[n],now;intBl[n],cnt[n];intNm, Q,tot,block;void Solve_2_5 () { for(intI=1; i<=n;i++) {ss[i]=ss[i-1]; sum[i]=sum[i-1];if((s[i]-' 0 ')%mod==0) sum[i]+=i,ss[i]++; } scanf ("%d", &q); for(intI=1; i<=q;i++) {intL,R;SCANF ("%d%d", &l,&r);printf("%lld\ n", sum[r]-sum[l-1]-(ss[r]-ss[l-1])*(L-1)); }}bool CMP (int x,int y){returnb[x]<b[y];} BOOL Cmp1 (W a,w b) {returnbl[a.l]<bl[b.l]| | (BL[A.L]==BL[B.L]&AMP;&AMP;A.R&LT;B.R);}intMain () {scanf ("%lld%s", &mod,s+1); N=strlen (s+1); block=sqrt(n);if(mod==2|| mod==5) {solve_2_5 ();return 0; } a[0]=1; for(intI=1; i<=n;i++) a[i]=Ten*a[I1]%mod; for(inti=n;i>=1; i--) b[i]= (A[n-i]*(s[i]-' 0 ') +b[i+1])%mod; for(intI=1; i<=n;i++) c[i]=i;Sort(c+1, c+n+1, CMP); for(intI=1; i<=n;i++)if(b[c[i]]==b[c[i-1]]) H[c[i]]=tot;ElseH[c[i]]=++tot; for(intI=1; i<=n;i++) bl[i]= (i-1)/block; scanf"%d", &q); for(intI=1; i<=q;i++) scanf ("%d%d",&Q[i].l,&Q[i]. R),Q[i]. p=i;Sort(Q+1,Q+q+1, CMP1);intL=1, r=0; for(intI=1; i<=q;i++) { while(l>Q[i]. L) now+=cnt[h[l-1]]+ (h[r+1]==h[l-1]), cnt[h[--l]]++; while(r<Q[i]. R) cnt[h[++r]]++,now+=cnt[h[r+1]]; while(l<Q[i]. L) Cnt[h[l]]--, now-=cnt[h[l]]+ (h[r+1]==h[l]), l++; while(r>Q[i]. R) now-=cnt[h[r+1]],cnt[h[r--]]--; ans[Q[i]. P]=now; } for(intI=1; i<=q;i++)printf("%lld\ n", Ans[i]);return 0;}

4542: [Hnoi2016] Large number | MO Team