HDU 6274 Master of Sequence (暴力+下整除)

Posted 2020-11-14 wrjlinkkkkkk

题意

两个1e5的数组a,b，定义(S(t)=left lfloor frac{t-b_i}{a_i} ight floor)，有三个操作
1 x y：将(a[x])变为(y)
2 x y：将(b[x])变为(y)
3 x：求使得(S(t)geq k)的最小(k)
其中(a_ileq 1000),(b_i,kleq 1e9)

思路

这题主要突破口在于(a_ileq 1000)
我们先从下整除下手，
(left lfloor frac{t-b_i}{a_i} ight floor=left lfloor frac{k_1a_i+c_1-k_2a_i-c_2}{a_i} ight floor=k_1-k_2+left lfloor frac{c_1-c_2}{a_i} ight floor)
其中
(c_1=tmod a_i)
(c_2=b_i mod a_i)
(c_1=t/ a_i)
(c_1=b_i/ a_i)
(left lfloor frac{c_1-c_2}{a_i} ight floor=left{egin{matrix} -1,c_1<c_2\ 0,c_1geq c_2 end{matrix} ight.)
那么
(S(t)=ret+sum_{i=1}^{1000}(frac{t}{i}cnt[i]-f[i][tmod i+1]))
其中
(f[x][y])表示(a_i=x)时，(y leq c_2 leq 1000)的个数
(cnt[x])表示(a_i=x)的个数，即(f[x][0])
(ret)表示所有数(k_2)的和
这题就做完了。。

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
    
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 998244353;
const int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int n,m;
ll a[maxn],b[maxn];
ll f[1111][1111];
ll ret;
ll S(ll t){
    ll ans = 0;
    for(int i = 1; i <= 1000; i++){
        ans+=t/i*f[i][0]-f[i][t%i+1];
    }
    return ans-ret;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= n; i++){
            scanf("%lld",&a[i]);
        }
        for(int i = 1; i <= n; i++){
            scanf("%lld", &b[i]);
        }
        mem(f,  0);
        ret = 0;
        for(int i = 1; i <= n; i++){
            ret+=b[i]/a[i];
            f[a[i]][b[i]%a[i]]++;
        }
        for(int i = 1; i <= 1000; i++){
            for(int j = i-1; j >= 0; j--){
                f[i][j]+=f[i][j+1];
            }
        }
        while(m--){
            int op,x;
            ll y;
            scanf("%d",&op);
            if(op<=2){
                scanf("%d %lld", &x ,&y);
                if(op==1){ 
                    ret-=b[x]/a[x];
                    ret+=b[x]/y;
                    for(int i = b[x]%a[x]; i >= 0; i--)f[a[x]][i]--;
                    for(int i = b[x]%y; i >= 0; i--)f[y][i]++;
                    a[x]=y;
                }
                else if(op==2){
                    ret-=b[x]/a[x];
                    ret+=y/a[x];
                    for(int i = b[x]%a[x]; i >= 0; i--)f[a[x]][i]--;
                    for(int i = y%a[x]; i >= 0; i--)f[a[x]][i]++;
                    b[x]=y;
                }
            }
            else{
                ll ans = 0;
                scanf("%lld", &y);
                ll l = 0,r = 1e13;
                while(l<=r){
                    ll mid = l+r>>1;
                    if(S(mid)>=y){
                        ans=mid;r=mid-1;
                    }
                    else l=mid+1;
                }
                printf("%lld
",ans);
            }

        }

    }
    return 0;
}
/*
2
4 6
2 4 6 8
1 3 5 7
1 2 3
2 3 3
3 15
1 3 8
3 90
3 66
8 5
2 4 8 3 1 3 6 24
2 2 39 28 85 25 98 35
3 67
3 28
3 73
3 724
3 7775
 */