CodeForces 616E

本文介绍了一种利用整数除法规律求解特定数学问题的方法,并给出了相应的C++代码实现。通过观察和总结规律,作者提出了一种高效算法,能够减少计算量并快速得出结果。

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打表发现一些规律,然后就可以推导一下就出来了,注意一下计算都是整数的整除。

n%i  和n%(n/i)是相等的,这个很好理解,而对于%的数从 n/i 到n/(i+1)+1 【逆序】, n%这些数是一个等差数列,而且差值显然是i。

于是就只要枚举到sqrt(n)就行了。注意m>n的时候的特判,以及细节处理,下标的特殊处理,每个地方都要%mod,mod下除法用费马小定理。

#include<cstdio>
#include<cstring>
#include<cmath>
#define mod 1000000007

long long n,m,ans;

long long min(long long a,long long b)
{
	if(a<b)
		return a;
	else
		return b;
}

long long qp(long long a,long long b)
{
	long long ans=1,cnt=a;
	while(b)
	{
		if(b&1)
			ans=(ans*cnt)%mod;
		cnt=(cnt*cnt)%mod;
		b>>=1;
	}
	return ans;
}

int main()
{
	scanf("%I64d%I64d",&n,&m);
	long long r=sqrt(n),st=0,ed=0,stnum,ednum,len,sum1;
	bool flag=false;
	if(r*r==n)
		r--;
	if(m>n)
		ans=(((m-n)%mod)*(n%mod))%mod;
	for(long long i=1;i<=r && i<=m;i++)
	{
		ans=(n%i+ans)%mod;
		if(flag)
			st=n/i,ed=n/(i+1)+1;
		else
			if(!flag && m>=(n/(i+1)+1))
				st=min(m,n),ed=n/(i+1)+1,flag=true;
		if(st)
		{
			stnum=(n%st);len=(st-ed+1);ednum=(stnum+(len-1)*i);
			len%=mod;
			sum1=(stnum+ednum)%mod;
			if(!(st==ed && st==i))
				ans=(ans+(((sum1*len)%mod)*qp(2,mod-2))%mod)%mod;
		}
	}
	printf("%I64d",ans);
	return 0;
}


### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update &#39;res&#39; for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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