bzoj4542: [Hnoi2016]大数

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=4542

思路：当P！=2或5时，显然10^x%P!=0

把后缀模P的值搞出来

于是问题就便成询问区间内%P为x的分别有多少个

这个再套一个莫队就可以了。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
const int maxn=100010;
typedef long long ll;
using namespace std;
struct quer{int l,r,id;}q[maxn];
ll ans[maxn],mod,a[maxn],n,m,pw[maxn],b[maxn],c[maxn],bel[maxn],tot,sz,cnt[maxn],now=0,scnt[maxn],ssum[maxn];char s[maxn];
bool cmp(quer a,quer b){return bel[a.l]!=bel[b.l]?bel[a.l]<bel[b.l]:a.r<b.r;}
bool back(quer a,quer b){return a.id<b.id;}

void init(){
	scanf("%lld%s%lld",&mod,s+1,&m),n=strlen(s+1),sz=(int)sqrt(n);
	pw[0]=1;for (int i=1;i<=n;i++) pw[i]=pw[i-1]*10%mod;
	for (int i=1;i<=n;i++) a[i]=s[i]-'0',bel[i]=(i-1)/sz+1;
	for (int i=n;i;i--) b[i]=(b[i+1]+a[i]*pw[n-i])%mod;//后缀和的模
	b[n+1]=0;
	memcpy(c,b,sizeof(c)),sort(c+1,c+2+n);
	tot=unique(c+1,c+2+n)-c-1;
	for (int i=1;i<=n+1;i++) b[i]=lower_bound(c+1,c+2+tot,b[i])-c;
	for (int i=1;i<=m;i++) scanf("%d%d",&q[i].l,&q[i].r),q[i].id=i,q[i].r++;
	
}

inline void modify(int p,int op){
	now-=cnt[b[p]]*(cnt[b[p]]-1)/2;
	cnt[b[p]]+=op;
	now+=cnt[b[p]]*(cnt[b[p]]-1)/2;
}

void work(){
	sort(q+1,q+1+m,cmp);
	for (int i=1,l=1,r=0;i<=m;i++){
		for (;r<q[i].r;r++) modify(r+1,1);
		for (;r>q[i].r;r--) modify(r, -1);
		for (;l<q[i].l;l++) modify(l, -1);
		for (;l>q[i].l;l--) modify(l-1,1);
		ans[q[i].id]=now;
	}
	for (int i=1;i<=m;i++) printf("%lld\n",ans[i]);
}

void spw(){
	for (int i=1;i<=n;i++) scnt[i]=scnt[i-1]+(a[i]%mod==0),ssum[i]=ssum[i-1]+(a[i]%mod==0?i:0);
	for (int i=1;i<=m;i++){
		int x=q[i].l,y=q[i].r-1;
		printf("%lld\n",ssum[y]-ssum[x-1]-1ll*(x-1)*(scnt[y]-scnt[x-1]));
	}
}

int main(){
	init();
	if (mod==2||mod==5) spw();else work();
	return 0;
}