The 2021 ICPC Asia Regionals Online Contest (II)

比赛链接

The 2021 ICPC Asia Regionals Online Contest (II)

G Limit

Description

Given $2n$ integers, $a_1,a_2 ,…,a_n ,b_1 ,b_2 ,…,b_n$ , and an integer t. You need to calculate:

$lim _{x→0}\frac{∑_{i=1}^na_i⋅ln(1+b_i⋅x)}{x^t}$.

Input

The first line consists of two integers n,t.

In the following n lines, the i-th line consists of two integers $a_i ,b_i$.

$1≤n≤100000,−100≤a_i,b_i≤100,0≤t≤5.$

Output

Please output the result of this limit. If the result is $∞$, please output "infinity" (without quotes). And if the result is an integer, please output this integer directly. Otherwise, the answer must be $b,a$ , such that a and b are coprime and $b≥2$, please output "a/b".

Sample Input 1

2 2
1 1
1 -1

Sample Output 1

-1

Sample Input 2

2 1
1 1
1 -1

Sample Output 2

0

Sample Input 3

2 3
1 1
1 -1

Sample Output 3

infinity

解题思路

洛必达法则

根据 $t$ 值，运用洛必达法则，即：分子为0时可以使用洛必达公式，否则答案为infinity，具体如下：t

• 时间复杂度：$O(nt)$

泰勒展开

直接对 $ln(1+b_i)$ 进行泰勒展开、变形即可，如下：

依上，对于 $j$，我们先枚举 $1\sim t-1$，如果第二重循环的和非0，即 $\frac{常数}{x^{t-j}}$，$x$ 趋于 $0$，则答案为infinity，否则 $j=t$ 时必有答案 $\frac{\sum_{i=1}^na[i]*pow(-b[i],t-1)*b[i]}{t}$

• 时间复杂度：$O(nt)$

代码

• 洛必达

//G:洛必达法则
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
using LL=long long;
int n,t,a[N],b[N];
LL res;
int main()
{
    scanf("%d%d",&n,&t);
    for(int i=1;i<=n;i++)
        scanf("%d%d",&a[i],&b[i]);
    if(t==0)puts("0");
    else
    {
        for(int i=1;i<=n;i++)
            res+=a[i]*b[i];
        if(t==1)
            printf("%lld\n",res);
        else if(res==0)
        {
            for(int i=1;i<=n;i++)
                res+=a[i]*b[i]*b[i];
            res/=(-2);
            if(t==2)
                printf("%lld\n",res);
            else if(res==0)
            {
                for(int i=1;i<=n;i++)
                    res+=a[i]*b[i]*b[i]*b[i];
                res/=3;
                if(t==3)
                    printf("%lld\n",res);
                else if(res==0)
                {
                    for(int i=1;i<=n;i++)
                        res+=a[i]*b[i]*b[i]*b[i]*b[i];
                    res/=4;
                    if(t==4)
                        printf("%lld\n",res);
                    else if(res==0)
                    {
                        for(int i=1;i<=n;i++)
                            res+=a[i]*b[i]*b[i]*b[i]*b[i]*b[i];
                        printf("%lld",res/30);
                    }
                    else
                        puts("infinity");
                }
                else
                    puts("infinity");
            }
            else
                puts("infinity");
        }
        else
            puts("infinity");
    }
    return 0;
}

• 泰勒公式

//G:泰勒展开
#include<bits/stdc++.h>
using namespace std;
using LL=long long;
const int N=1e5+10;
int n,t,a[N],b[N];
LL res;
int main()
{
    scanf("%d%d",&n,&t);
    for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]);
    if(!t)
        puts("0");
    else
    {
        for(int i=1;i<t;i++)
        {
            res=0;
            for(int j=1;j<=n;j++)
                res+=a[j]*pow(-b[j],i-1)*b[j];
            if(res)
            {
                puts("infinity");
                return 0;
            }
        }
        for(int i=1;i<=n;i++)res+=a[i]*pow(-b[i],t-1)*b[i];
        res/=t;
        printf("%lld\n",res);
    }
    return 0;
}

L Euler Function

Description

The Euler function, $φ$, is known to be important in math theory. Given a positive integer $n$, $φ(n)$ is defined as the number of integers in $[1,n]$ that are co-prime with $n$. For example, $φ(1)=1,φ(10)=4,φ(11)=10$.

Now, given an array $x$ with $n$ positive integers $x_1,…,x_n, your task is to maintain $m$ operations on $x$, which can be categorized into two types:

• $0\,l\, r\, w$, for each index $i∈[l,r]$, change x_i to $x_i\times w$.

• $1\, l\, r$, calculate and print$(∑_{i=l}^rφ(x_i))\bmod998244353$.

Input

The first line contains two inetegers $n,m (1≤n,m≤10^5)$.

The second line contains $n$ integers $x_1,…,x_n(1≤x_i≤100)$.

Then $m$ lines follow. The $i$-th line describes the $i$-th operation. The input guarantees that $1≤l≤r≤n$ and $1≤w≤100$.

Output

For each operation of type $1$, output a single line with a single integer, representing the answer.

Sample Input

5 5
1 2 3 4 5
1 1 5
0 1 3 2
1 1 5
0 2 5 6
1 1 5

Sample Output

10
11
37

解题思路

• 时间复杂度：$O()$

代码

M Addition

Description

The numbers in the computer are usually stored in binary form. But in Little A's computer, the numbers are stored in a different way.

The computer uses n bits to store a number. The value it stores is $∑_{i=0}^{n−1}v_isgn_i2^i$
, where $v$ is an array of length $n$ containing only $0$ and $1$, and $sgn$ is a predefined array of length $n$ containing only $−1$ and $1$. It is not difficult to find that every expressible integer has a unique expression.

Little A gives you the binary representation of $a$ and $b$ in his computer, and you should report the binary representation of $a+b$ in his computer. It is guaranteed that $max{∣a∣,∣b∣}≤10 ^8$
, and all integers in $[−10^9,10^9]$ can be expressed in his computer.

Input

The first line contains an integer $n$, which represents the number of bits used to store an integer.

The second line contains $n$ integers, and the $i$-th of them represents $sgn_{i−1}$ .

The third line contains $n$ integers, and the $i$-th of them represents $va_{i−1}$ . The value of a is $∑_{i=0}^{n−1}va_isgn_i2^i$ .

The fourth line contains $n$ integers, and the $i$-th of them represents $vb_{i−1}. The value of $b$ is $∑_{i=0}^{n−1}vb_isgn_i2^i$.

It is guaranteed that $32≤n≤60,sgn_i∈{−1,1},va _i,vb_i ∈{0,1}$, and $max{∣a∣,∣b∣}≤10^8$.

Output

Output one line containing $n$ integers, separated by spaces. The $i$-th of them represents $vc_{i−1}∈{0,1}$. There should be no extra spaces at the end of the line.

You should guarantee that $a+b=∑_{i=0}^{n−1}vc_isgn_i2^i$.

Your output must be in a single line.

Sample Input

32
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -1
1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Sample Output

0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

解题思路

模拟二进制加法

与二进制不同的是，本题多了每位的符号位，关键在于处理进位，当当前位与后一位符号位相同时，后一位加上的正是进位，否则是进位的相反数，由于一开始的进位 $carry=0$，对下一位的进位只能是 $0\sim 1 \times sgn[2]$，即：$-1\sim 1$，故:$a[i]+b[i]+carry$ 最小是 $-1$ （即需借位），最大是 $3$ ,分别讨论即可~

• 时间复杂度：$O(n)$

代码

#include<bits/stdc++.h>
using namespace std;
const int N=65;
int a[N],b[N],c[N],sgn[N],n;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&sgn[i]);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)scanf("%d",&b[i]);
    int carry=0;
    for(int i=1;i<=n;i++)
    {
        carry*=sgn[i];
        int t=a[i]+b[i]+carry;
        if(t==0||t==1)
        {
            c[i]=t;
            carry=0;
        }
        if(t==2)
        {
            c[i]=0;
            carry=1;
        }
        if(t==3)
        {
            c[i]=1;
            carry=1;
        }
        if(t==-1)
        {
            c[i]=1;
            carry=-1;
        }
        carry*=sgn[i];
    }
    for(int i=1;i<=n;i++)printf("%d%c",c[i]," "[i==n]);
    return 0;
}

__EOF__

本文作者：acwing_zyy
本文链接：https://www.cnblogs.com/zyyun/p/15355314.html
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