Codeforces Round #476 (Div. 2) [Thanks, Telegram!] C

http://codeforces.com/contest/965/problem/C

题目大意:n个糖,k个人,每次最多只能拿M个糖,从第一个人开始拿,可以循环D次。问Arkady最多可以拿几块糖?Arkady 是第一个人

思路:

设每次拿x个糖,循环i次,所以贪心就是x*(d-1)*k + x <= n

求出x就好了

爆LL是真的烦

//看看会不会爆int!数组会不会少了一维!
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n")
LL n, k, m, d;


int main(){
    cin >> n >> k >> m >> d;
    LL ans = min(n, m);
    for (LL i = 1; i <= d; i++){
        LL tmp = (i-1) * k + 1;
        if (tmp <= 0) continue;
        LL x = n / tmp;
        if (x > m) x = m;
        if (x * (i - 1) > (n - x) / k) continue;
        LL res = x * (i-1) + x;
        if (res > ans){
            ans = res;
        }
    }
    cout << ans <<endl;
    return 0;
}

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转载自www.cnblogs.com/heimao5027/p/8971644.html