A. K-divisible Sum

A. K-divisible Sum

思路

\[ans = \left \lceil \frac{kx}{n} \right \rceil \]

\[x = x_{min} \ge \left \lceil \frac{n}{k} \right \rceil \]

代码

点击查看代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<cctype>
using namespace std;

#define X first
#define Y second

typedef pair<int,int> pii;
typedef long long LL;
const char nl = '\n';
const int N = 1e5+10;
const int M = 2e5+10;
int n,k;

bool check(int x){
	return x >= ceil(1.0 * n / k);
}

void solve(){
	int t;
	cin >> t;
	while(t -- ){
		cin >> n >> k;
		int l = 1,r = 1e9;
		while(l < r){
			int mid = l + r >> 1;
			if(check(mid))r = mid;
			else l = mid + 1;
		}
		cout << (int)ceil((long double)k * l / n) << '\n';	
	} 
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);

	solve();
}

注意

1. \(k * l\)会爆int需要先转为long double
2. 1e+09 != 100000000 输出时还要转回int

技巧

1. 向上取整转化为向下取整【c++默认向下取整】

\[\left \lceil \frac{n}{k} \right \rceil = \left \lfloor \frac{n + k - 1}{k} \right \rfloor \]

这样可以避免类型转换带来不必要的错误