codeforces822D(思维)

最新推荐文章于 2022-09-14 11:11:13 发布

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本文介绍了一种针对多人参与竞赛时实现最小平局分配次数的算法。通过预先计算每个可能人数对应的平局次数并利用动态规划技巧，快速得出在特定人数范围内达到最少平局所需的总次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

思路：对于n个人进行平局分配次数最少就是每局的人最少即每次分组越多越好。

打个表，就行。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef set<int>::iterator ITER;
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define clr(x,y) memset(x,y,sizeof x)

const int maxn = 5e6 + 10;
const int Mod = 1e9 + 7;
const int N = 2;
struct Matrix{Matrix(){clr(m,0);}int m[N][N];};

Matrix Mul(Matrix a,Matrix b){Matrix c;for(int i = 0; i < N; i ++){for(int k = 0; k < N; k ++){if(a.m[i][k] == 0)continue;for(int j = 0; j < N; j ++)c.m[i][j] += a.m[i][k] * b.m[k][j] % Mod,c.m[i][j] %= Mod;}}return c;}
ll Mul(ll a,ll b){a %= Mod;b %= Mod;ll ret = 0;while(b){if(b & 1){ret += a;if(ret > Mod)ret -= Mod;}b >>= 1;a = (a << 1) % Mod;}return ret;}
Matrix pows(Matrix x,ll n){Matrix ret;for(int i = 0; i < 2; i ++)ret.m[i][i] = 1;while(n){if(n & 1)ret = Mul(ret,x);x = Mul(x,x);n >>= 1;}return ret;}
ll pows(ll x,ll n){ll ret = 1;while(n){if(n & 1)ret = ret * x % Mod;x = x * x % Mod;n >>= 1;}return ret;}
ll powss(ll x,ll n){ll ret = 1;while(n){if(n & 1)ret = Mul(ret,x);x = Mul(x,x);n >>= 1;}return ret;}
ll gcd(ll x,ll y){return y ? gcd(y,x % y):x;}
ll euler(int n){int ret = n;for(int i = 2; i * i<= n; i ++)if(n % i == 0){ret = ret / i * (i - 1);while(n % i == 0)n /= i;}if(n > 1)ret = ret / n * (n - 1);return ret;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d = a;x = 1;y = 0;return;}ex_gcd(b,a % b,d ,y,x);y -=a/b * x;}
inline int lowbit(int x){return x &(-x);}

//void update(int x,int val){for(int i = x; x < maxn; i += lowbit(i))tree[i] += val;}
//void get(int x){int ret = 0;for(int i = x; i > 0 ;i -= lowbit(i))ret += tree[i];return ret;}
//int finds(int x){return fa[x] == x ? x : (fa[x] = finds(fa[x]));}

ll dp[maxn];
bool is_[maxn];
int prime[maxn];
void Init()
{
	clr(is_,true);
	clr(dp,0);
	is_[0] = is_[1] = false;dp[1] = 0;
	int len = 0;
	for(int i = 2; i < maxn; i ++)
	{
		if(is_[i])
		{
			prime[len ++] = i;
			dp[i] = 1ll * i * (i - 1)/2;
			for(int j = i * 2; j < maxn;j += i)
			{
				is_[j] = false;
			}
		}
	}
	
	for(int i = 2;i < maxn;i ++)
	{
		if(dp[i])continue;
		for(int j = 0; j < len && prime[j] < i; j ++)
		{
			if(i % prime[j] == 0)
			{
				dp[i] = (i/prime[j] * dp[prime[j]] % Mod+ dp[i/prime[j]]) % Mod;break;
			}
		}
	}
}
int main()
{
	Init();
	ll t,l,r;
	while( ~ scanf("%I64d%I64d%I64d",&t,&l,&r))
	{
		
		ll ans = 0;
		for(int i = r; i >= l;i --)
		{
			ans = ans* t % Mod;
			ans = (ans + dp[i])% Mod;
		}
		
		printf("%I64d\n",ans);
	}
	return 0;
}
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