2017 ACM-ICPC 亚洲区(西安赛区)网络赛 B Coin (概率计算)
2017-10-05 13:59
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传送门: https://nanti.jisuanke.com/t/17115
Bob has a not even coin, every time he tosses the coin, the probability that the coin's front face up is \frac{q}{p}(\frac{q}{p}
\le \frac{1}{2})pq(pq≤21).
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 \frac{X}{Y}YX,
because the answer could be extremely large, you only need to print (X
* Y^{-1}) \mod (10^9+7)(X∗Y−1)mod(109+7).
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≤107) indicates
the i-th test case.
For each test case, print an integer in a single line indicates the answer.
2017
ACM-ICPC 亚洲区(西安赛区)网络赛
题目的意思很简单,就是让你扔硬币,给你p,q.向上的概率为q/p;然后问
你向上次数为偶数的概率。
这个地方用到高中组合知识:(组合数学,概率方面的问题)
设 向上概率概率是 a 向下为b 则 a+b =1;
k次 扔 取时 向上为偶数的为: C(k,0) *a^0 * b^k + C(k,2) *a^2 *b^(k-
2) +C(k,4)*a^4 *b^(k-4) ......... + C(k,k)*a^k*b^0
(a+b)^k 展开为 : C(k,0)*a^0 *b^k,+ C(k,1)*a^1*b^(k-1)
+C(k,2)*a^2*b^(k-2) .......+C(k,k)*a^k*b^0
(a-b) ^k 展开为: C(k,0) *a^0 *(-b)^k +C(k,1)^a* (-b)^(k-1)
.........+C(k,k)*a^k*(-b^0)
所以 应为: ((a+b)^k +(a-b)^k ) /2
题意 要求逆元:
求逆元方法;:(三种)
ll inv_exgcd(ll a,ll n){lld,x,y;ex_gcd(a,n,d,x,y);return d==1?
(x+n)%n:-1;}
ll inv1(ll b){returnb==1?1:(MOD-MOD/b)*inv1(MOD%
b)%MOD;}
ll inv2(ll b){return qpow(b,MOD-2);}
求逆元 应用 费马小定理的那个 即为:
ll inv2(ll b){return qpow(b,MOD-2);}
若用 带/ 法的 求逆元 则会出现 /0 的情况 WA
#include<stdio.h>
#include<iostream>
#include <algorithm>
#include<string.h>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#define LL long long
#define INf 0x3f3f3f3f
const int MOD=1e9+7;
using namespace std;
LL qpow(LL x,LL n)
{
LL res=1;
for(;n;n>>=1)
{
if(n&1)
res=(res*x)%MOD;
x=(x*x)%MOD;
}
return res;
}
LL inv2(LL b)
{
return qpow(b,MOD-2);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
LL p,q,k;
scanf("%lld %lld %lld",&p,&q,&k);
LL a=qpow(p,k);
LL sum=(a+qpow((p-2*q),k))%MOD;;
sum=sum*(inv2(2*a)%MOD);
printf("%lld\n",sum%MOD);
}
return 0;
}
Bob has a not even coin, every time he tosses the coin, the probability that the coin's front face up is \frac{q}{p}(\frac{q}{p}
\le \frac{1}{2})pq(pq≤21).
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 \frac{X}{Y}YX,
because the answer could be extremely large, you only need to print (X
* Y^{-1}) \mod (10^9+7)(X∗Y−1)mod(109+7).
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≤107) 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
2017ACM-ICPC 亚洲区(西安赛区)网络赛
题目的意思很简单,就是让你扔硬币,给你p,q.向上的概率为q/p;然后问
你向上次数为偶数的概率。
这个地方用到高中组合知识:(组合数学,概率方面的问题)
设 向上概率概率是 a 向下为b 则 a+b =1;
k次 扔 取时 向上为偶数的为: C(k,0) *a^0 * b^k + C(k,2) *a^2 *b^(k-
2) +C(k,4)*a^4 *b^(k-4) ......... + C(k,k)*a^k*b^0
(a+b)^k 展开为 : C(k,0)*a^0 *b^k,+ C(k,1)*a^1*b^(k-1)
+C(k,2)*a^2*b^(k-2) .......+C(k,k)*a^k*b^0
(a-b) ^k 展开为: C(k,0) *a^0 *(-b)^k +C(k,1)^a* (-b)^(k-1)
.........+C(k,k)*a^k*(-b^0)
所以 应为: ((a+b)^k +(a-b)^k ) /2
题意 要求逆元:
求逆元方法;:(三种)
ll inv_exgcd(ll a,ll n){lld,x,y;ex_gcd(a,n,d,x,y);return d==1?
(x+n)%n:-1;}
ll inv1(ll b){returnb==1?1:(MOD-MOD/b)*inv1(MOD%
b)%MOD;}
ll inv2(ll b){return qpow(b,MOD-2);}
求逆元 应用 费马小定理的那个 即为:
ll inv2(ll b){return qpow(b,MOD-2);}
若用 带/ 法的 求逆元 则会出现 /0 的情况 WA
#include<stdio.h>
#include<iostream>
#include <algorithm>
#include<string.h>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#define LL long long
#define INf 0x3f3f3f3f
const int MOD=1e9+7;
using namespace std;
LL qpow(LL x,LL n)
{
LL res=1;
for(;n;n>>=1)
{
if(n&1)
res=(res*x)%MOD;
x=(x*x)%MOD;
}
return res;
}
LL inv2(LL b)
{
return qpow(b,MOD-2);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
LL p,q,k;
scanf("%lld %lld %lld",&p,&q,&k);
LL a=qpow(p,k);
LL sum=(a+qpow((p-2*q),k))%MOD;;
sum=sum*(inv2(2*a)%MOD);
printf("%lld\n",sum%MOD);
}
return 0;
}
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