2017 ACM-ICPC 亚洲区（西安赛区）网络赛

最新推荐文章于 2024-05-13 15:08:27 发布

feng_zhiyu

最新推荐文章于 2024-05-13 15:08:27 发布

阅读量723

点赞数

分类专栏：题集文章标签： acm-icpc 网络赛

本文链接：https://blog.youkuaiyun.com/feng_zhiyu/article/details/78008546

版权

题集专栏收录该内容

20 篇文章

订阅专栏

本文解析了两个竞赛题目：B.Coin通过组合数和快速幂计算掷硬币出现正面偶数次的概率；C.Sum寻找使得特定乘积后的数字和能被233整除的正整数k。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

B. Coin（组合数+快速幂，费马小求逆元）
1000ms 32768K

Bob has a not even coin, every time he tosses the coin, the probability that the coin’s front face up is qp (qp≤12) 

The question is, when Bob tosses the coin kk times, what’s the probability that the frequency of the coin facing up is even number.

If the answer is XY , because the answer could be extremely large, you only need to print(X∗Y−1)mod(109+7).
Input Format

First line an integer T, indicates the number of test cases (T≤100).

Then Each line has 3 integer p,q,k(1≤p,q,k≤107) indicates the i-th test case.

Input Format

First line an integer TT, indicates the number of test cases (T \le 100T≤100).

Then Each line has 33 integer p,q,k(1\le p,q,k \le 10^7)p,q,k(1≤p,q,k≤10
7
 ) indicates the i-th test case.

Output Format

For each test case, print an integer in a single line indicates the answer.

样例输入

2
2 1 1
3 1 2
样例输出

500000004
555555560
题目来源

2017 ACM-ICPC 亚洲区（西安赛区）网络赛

题意：（可能我英文菜，但是确定我每个地方都知道。。。）
给你一个不均匀的硬币，掷一次出现正面的概率是q/p，掷硬币k次出现正面偶数次的概率和为X/Y，输出((X*Y^(-1))%(1e9+7).

分析：
这里Y^(-1)就用逆元求就是了
这里我理解错了，偶数次，把0给漏了。。
求概率和转化为次数为求出现正面偶数次的情况，然后就是组合数公式的推理
看过程点击http://blog.youkuaiyun.com/winter2121/article/details/78005036?locationNum=10&fps=1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <set>
#include <cmath>
#include <string>
using namespace std;
#define mem(a,n) memset(a,n,sizeof(a))
typedef long long LL;
const double eps=1e-6;
const int N=1e3+5;
const int mod=1e9+7;
LL pow_mod(LL a,LL b)
{
    LL ans=1;
    a%=mod;
    while(b>0)
    {
        if(b&1) ans=(ans*a)%mod;
        b>>=1;
        a=(a*a)%mod;
    }
    return ans;
}
LL fermat(LL a,LL p)///费马小定理求逆元
{
    return pow_mod(a,p-2);
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        LL p,q,k;
        cin>>p>>q>>k;
        // printf("%d  %d\n",p,k);
        LL tmp=(pow_mod(p,k)+pow_mod(p-2*q,k))%mod*fermat(2,mod)%mod;
        LL tt=(LL)pow_mod(p,k);
        //printf("tt=%lld tmp=%lld\n",tt,tmp);
        cout<<(tmp*fermat(tt,mod))%mod<<endl;
    }
    return 0;
}

C. Sum
1000ms 32768K
Define the function S(x)S(x) for xx is a positive integer. S(x)S(x) equals to the sum of all digit of the decimal expression of xx. Please find a positive integer kk that S(k*x)\%233=0S(k∗x)%233=0.

Input Format

First line an integer TT, indicates the number of test cases (T \le 100T≤100). Then Each line has a single integer x(1 \le x \le 1000000)x(1≤x≤1000000) indicates i-th test case.

Output Format

For each test case, print an integer in a single line indicates the answer. The length of the answer should not exceed 20002000. If there are more than one answer, output anyone is ok.

样例输入

1
1
样例输出

89999999999999999999999999
题目来源

2017 ACM-ICPC 亚洲区（西安赛区）网络赛

分析：任意数乘9得到的数的数位和仍然是9的倍数，直接输出233个9就可以了