洛谷P3245 [HNOI2016]大数 莫队

Posted 2020-11-02 Mychael

题目

题解

除了\(5\)和\(2\)
后缀数字对\(P\)取模意义下，两个位置相减如果为\(0\)，那么对应子串即为\(P\)的倍数
只用对区间种相同数个数\(x\)贡献\({x \choose 2}\)
经典莫队题
\(P = 2\)或\(5\)就特判一下

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<‘ ‘; puts("");
using namespace std;
const int maxn = 100005,maxm = 100005,INF = 1000000000;
LL n,m,P,B;
char s[maxn];
LL ans[maxn];
struct Que{int l,r,b,id;}q[maxn];
inline bool operator <(const Que& a,const Que& b){
    return a.b == b.b ? a.r < b.r : a.l < b.l;
}
void solve1(){
    scanf("%lld",&m);
    REP(i,m) scanf("%d%d",&q[i].l,&q[i].r),q[i].id = i,q[i].b = q[i].l / B;
    sort(q + 1,q + 1 + m);
    LL L = q[1].l,R = q[1].r; LL cnt = 0,sum = 0;
    for (int i = L; i <= R; i++){
        if ((s[i] - ‘0‘) % P == 0) cnt++,sum += i - L + 1;
    }
    ans[q[1].id] = sum;
    for (int i = 2; i <= m; i++){
        while (L != q[i].l || R != q[i].r){
            if (L < q[i].l){
                sum -= cnt;
                if ((s[L] - ‘0‘) % P == 0) cnt--;
                L++;
            }
            if (L > q[i].l){
                L--;
                if ((s[L] - ‘0‘) % P == 0) cnt++;
                sum += cnt;
            }
            if (R < q[i].r){
                R++;
                if ((s[R] - ‘0‘) % P == 0) cnt++,sum += R - L + 1;
            }
            if (R > q[i].r){
                if ((s[R] - ‘0‘) % P == 0) cnt--,sum -= R - L + 1;
                R--;
            }
        }
        ans[q[i].id] = sum;
    }
    REP(i,m) printf("%lld\n",ans[i]);
}
int b[maxn],bi,a[maxn],tot,bac[maxn];
int getn(int x){return lower_bound(b + 1,b + 1 + tot,x) - b;}
LL C(LL x){
    if (x <= 1) return 0;
    return x * (x - 1) / 2;
}
void solve2(){
    for (int i = n,bin = 1; i; i--,bin = bin * 10 % P){
        b[i] = a[i] = ((s[i] - ‘0‘) * bin % P + a[i + 1]) % P;
    }
    n++;
    sort(b + 1,b + 1 + n); tot = 1;
    for (int i = 2; i <= n; i++) if (b[i] != b[tot]) b[++tot] = b[i];
    for (int i = 1; i <= n; i++) a[i] = getn(a[i]);
    scanf("%lld",&m);
    REP(i,m){
        scanf("%d%d",&q[i].l,&q[i].r); q[i].r++;
        q[i].id = i,q[i].b = q[i].l / B;
    }
    sort(q + 1,q + 1 + m);
    LL L = q[1].l,R = q[1].r; LL sum = 0;
    for (int i = L; i <= R; i++){
        sum -= C(bac[a[i]]);
        sum += C(++bac[a[i]]);
    }
    ans[q[1].id] = sum;
    for (int i = 2; i <= m; i++){
        while (L != q[i].l || R != q[i].r){
            if (L < q[i].l){
                sum -= C(bac[a[L]]);
                sum += C(--bac[a[L]]);
                L++;
            }
            if (L > q[i].l){
                L--;
                sum -= C(bac[a[L]]);
                sum += C(++bac[a[L]]);
            }
            if (R < q[i].r){
                R++;
                sum -= C(bac[a[R]]);
                sum += C(++bac[a[R]]);
            }
            if (R > q[i].r){
                sum -= C(bac[a[R]]);
                sum += C(--bac[a[R]]);
                R--;
            }
        }
        ans[q[i].id] = sum;
    }
    REP(i,m) printf("%lld\n",ans[i]);
}
int main(){
    scanf("%lld%s",&P,s + 1);
    n = strlen(s + 1); B = (int)sqrt(n) + 1;
    if (P == 2 || P == 5) solve1();
    else solve2();
    return 0;
}