代碼用時：1h

10W級的數據跑莫隊比較正常吧。註意特判P==2 || P==5的情況（只需要判斷個位數即可）。

WA了兩次，struct D中x變量應該是long long的。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(i,l,r) for (int i=l; i<=r; i++)
typedef long long ll;
using namespace std;

const int N=200100;
int n,m,x,y,cnt,bl[N],num[N],c[N];
ll P,ans,res[N];
 
char S[N];
struct D{ ll x; int y; }s[N];
struct T{ int l,r,id; }a[N];

bool cmp(D a,D b){ return a.x<b.x; }
bool cmp1(T a,T b){ return (bl[a.l]==bl[b.l]) ? a.r<b.r : bl[a.l]<bl[b.l]; }
void work(int x,int k){
    if (k==1) ans+=num[c[x]],num[c[x]]++;
            else num[c[x]]--,ans-=num[c[x]];
}

 
void solve(){
    rep(i,1,n){
        s[i]=s[i-1];
        if (!((S[i]-‘0‘)%P)) s[i].x+=i,s[i].y++;
    }
    while (m--) scanf("%d%d",&x,&y),printf("%lld\n",s[y].x-s[x-1].x-1ll*(x-1)*(s[y].y-s[x-1].y));
}

int main(){
    freopen("bzoj4542.in","r",stdin);
    freopen("bzoj4542.out","w",stdout);
    scanf(
 
"%lld",&P); scanf("%s%d",S+1,&m);
    n=strlen(S+1); ll now=1;
    if (P==2 || P==5) { solve(); return 0; }
    rep(i,1,m) scanf("%d%d",&a[i].l,&a[i].r),a[i].r++,a[i].id=i;
    for (int i=n; i; i--,now=now*10%P) s[i].x=(s[i+1].x+(S[i]-‘0‘)*now)%P,s[i].y=i;
    s[++n].x=0; s[n].y=n; sort(s+1,s+n+1,cmp);
    rep(i,1,n){
        if (i==1 || s[i].x!=s[i-1].x) cnt++;
        c[s[i].y]=cnt;
    }
    int sz=(int)sqrt(n)+1;
    rep(i,1,n) bl[i]=(i-1)/sz;
    sort(a+1,a+m+1,cmp1);
    int L=1,R=1; num[c[1]]++;
    rep(i,1,m){
         while (R<a[i].r) work(++R,1);
         while (L>a[i].l) work(--L,1);
         while (R>a[i].r) work(R--,-1);
         while (L<a[i].l) work(L++,-1);
         res[a[i].id]=ans;
    }
    rep(i,1,m) printf("%lld\n",res[i]);
    return 0;
}

[BZOJ4542][HNOI2016]大數(莫隊)