CodeForces -1168A Increasing by Modulo（二分答案）

题目链接：点击这里

题目大意：
给出 n , m n,m n,m 和一个长度为 n n n 的序列 a 1 , a 2 , . . . , a n a_1,a_2,...,a_n a1​,a2​,...,an​
每次操作可以选择一个数 k k k 和 k k k 个数 b 1 , b 2 , . . . , b k b_1,b_2,...,b_k b1​,b2​,...,bk​ ，使得 a b i = ( a b i + 1 ) m o d    m a_{b_i}=(a_{b_i}+1) \mod m abi​​=(abi​​+1)modm ，求使序列 a a a 单调不下降的最小操作次数

题目分析：
最小操作次数显然满足答案的单调性，我们考虑二分答案的 c h e c k check check 如何写：我们对相邻元素 a [ i ] , a [ i − 1 ] a[i],a[i-1] a[i],a[i−1] 的大小关系分三种情况讨论：

1. a [ i ] = a [ i − 1 ] a[i]=a[i-1] a[i]=a[i−1] 此时让 a [ i ] a[i] a[i] 保持不变即可
2. a [ i ] > a [ i − 1 ] a[i]>a[i-1] a[i]>a[i−1] 如果能让 a [ i ] a[i] a[i] 通过循环变成 a [ i − 1 ] a[i-1] a[i−1] 更好，此时需要操作次数为 a [ i − 1 ] + m − a [ i ] a[i-1]+m-a[i] a[i−1]+m−a[i] ，如果操作次数不够就维持原样即可
3. a [ i ] < a [ i − 1 ] a[i]<a[i-1] a[i]<a[i−1] 此时我们需要让 a [ i ] a[i] a[i] 增加到 a [ i − 1 ] a[i-1] a[i−1] 使之保持序列的单调不下降，需要的操作次数为 a [ i − 1 ] − a [ i ] a[i-1]-a[i] a[i−1]−a[i]

具体细节见代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<set>
#include<map>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
ll read()
{
	ll res = 0,flag = 1;
	char ch = getchar();
	while(ch<'0' || ch>'9')
	{
		if(ch == '-') flag = -1;
		ch = getchar();
	}
	while(ch>='0' && ch<='9')
	{
		res = (res<<3)+(res<<1)+(ch^48);//res*10+ch-'0';
		ch = getchar();
	}
	return res*flag;
}
const int maxn = 5e5+5;
const int mod = 1e9+7;
const double pi = acos(-1);
const double eps = 1e-8;
ll n,m,a[maxn],tmp[maxn];
bool check(int val)
{
	memcpy(tmp,a,sizeof(a));
	for(int i = 1;i <= n;i++)
		if(tmp[i] < tmp[i-1])
		{
			if(tmp[i-1]-tmp[i] > val) return false;
			tmp[i] = tmp[i-1];
		} 
		else if(tmp[i-1]+m-tmp[i] <= val) tmp[i] = tmp[i-1];
	return true;
}
int main()
{
	n = read(),m = read();
	for(int i = 1;i <= n;i++)
		a[i] = read();
	int l = 0,r = m,ans = m;
	while(l <= r)
	{
		int mid = l+r>>1;
		if(check(mid))
		{
			ans = min(ans,mid);
			r = mid-1;
		}
		else l = mid+1;
	}
	printf("%d\n",ans);
	return 0;
}

标签：tmp,...,ch,Modulo,Increasing,ll,CodeForces,mid,include
来源： https://blog.csdn.net/qq_39641976/article/details/111566461