F. Final Boss

合集 - 二分(7)

1.F. Rudolf and Imbalance2024-03-142.D. Blocking Elements2024-03-153.P8736 [蓝桥杯 2020 国 B] 游园安排2024-03-304.C. Permutation Counting2024-05-14

5.F. Final Boss2024-06-13

6.C. To Become Max2024-07-177.P10838 『FLA - I』庭中有奇树2024-08-07

收起

原题链接

题解

1.由于一回合可以使用多次技能，所以直接二分回合数即可
2.回合数最多为 $4^{10}$

code

#include<bits/stdc++.h>
using namespace std;

#define ll long long

ll a[200005],c[200005];
ll h,n;

inline void read(ll &x) {
    x = 0;
    ll flag = 1;
    char c = getchar();
    while(c < '0' || c > '9'){
        if(c == '-') flag = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + (c ^ 48);
        c = getchar();
    }
    x *= flag;
}

inline void write(ll x)
{
    if(x < 0){
        putchar('-');
        x = -x;
    }
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}

bool check(ll len)
{
    ll ans = 0;
    for(ll i = 1; i <= n; i++)
    {
        ans += a[i] * ((len-1LL) / c[i]+1LL);
        if(ans>=h) return 1;
    }
    return 0;
}

int main()
{
    ll t;
    read(t);
    while(t--)
    {
        read(h);
        read(n);

        for(ll i = 1; i <= n; i++) read(a[i]);
        for(ll i = 1; i <= n; i++) read(c[i]);

        ll l = 0, r = 4e11;
        while(l + 1LL < r)//时间复杂度最多为40*n
        {
            ll mid = (l + r) / 2LL;
            if(check(mid)) r = mid;
            else l = mid;
        }

        write(r);
        putchar('\n');
    }
    return 0;
}