RMRC2016: B Election 组合数学

B: Election

时间限制: 1 Sec   内存限制: 128 MB
提交: 122   解决: 21
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题目描述

After all the fundraising, campaigning and debating, the election day has finally arrived. Only two candidates remain on the ballot and you work as an aide to one of them.
Reports from the voting stations are starting to trickle in and you hope that you can soon declare a victory.
There are N voters and everyone votes for one of the two candidates (there are no spoiled ballots). In order to win, a candidate needs more than half of the votes. A certain number M≤N of ballots have been counted, and there are Vi votes for candidate i (V1+V2 = M), where V1 is the number of votes your candidate received.
Due to the historical data and results of highly scientific polls, you know that each of the remaining votes has a 50% chance to go to your candidate. That makes you think that you could announce the win before all the votes are counted. So, if the probability of winning strictly exceeds a certain threshold W, the victory is yours! We just hope you are sure of this, we don’t want any scandals...

输入

The first line of input contains a single positive integer T≤100 indicating the number of test cases. Next
T lines each contain four integers: N, V1, V2 and W as described above.
The input limits are as follows:
1≤N≤50
50≤W<100
V1,V2≥0
V1 + V2≤N

输出

For each test case print a single line containing the appropriate action:
If the probability that your candidate will win is strictly greater than W%, print
         GET A CRATE OF CHAMPAGNE FROM THE BASEMENT!
If your candidate has no chance of winning, print
         RECOUNT!
Otherwise, print
         PATIENCE, EVERYONE!

样例输入

4

5 0 3 75

5 0 2 75

6 1 0 50

7 4 0 75

样例输出

RECOUNT!

PATIENCE, EVERYONE!

PATIENCE, EVERYONE!

GET A CRATE OF CHAMPAGNE FROM THE BASEMENT!

提示

自问 再也没有比这道题 恶心的题了;

题意 真心 mmp;

题意: 投票选举  有n 现在有 v1 对方有v2  剩下票 每一张 获得的概率为1/2;  获胜 必须 票数 超过 n/2

问你 当前状态下 推算  你获胜的概率是不是 大于w  是 yes  如果  剩下票数全部给你 也不够 n/2  则 no

其余 不超过w  输出possible;

现在有v1 张  剩下 票数为 t1:  n-v1-v2;  离获胜n/2  还缺 t2:  n/2 -v1+1  考虑到n为奇数的问题 所以+1

从  t1 要至少选择t2 张   

C (t1,t2) *  1/2 ^t2 * 1/2 ^ t1-t2     =C(t1,t2) * 1/2^t1    这个意思是 1/2 获胜概率 1/2 失败概率   获胜是 t2  失败 是 t1-t2 所以总是为  t1

Sigma  C(t1,t2)---C(t1,t1)  * 1/2 ^t1   >  w  则 获胜   转换为  Sigma  C(t1,t2)---C(t1,t1)  >  2^t1 * w

 失败 是  v1 + n-v1-v2 <= n/2   化简为 2*v2 >=n 

注意  在求组合公式 和 快速幂时  不需要 不需要 不需要 mod      因为 %mod 了 所以 mmp 一直wa  

代码:

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <vector>
#define mem(a,b) memset(a,b,sizeof(a))
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define FIN      freopen("input.txt","r",stdin)
#define FOUT     freopen("output.txt","w",stdout)
#define S1(n)    scanf("%d",&n)
#define SL1(n)   scanf("%I64d",&n)
#define S2(n,m)  scanf("%d%d",&n,&m)
#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)
#define Pr(n)     printf("%d\n",n)

using namespace std;
typedef long long ll;
const double PI=acos(-1);
const int INF=0x3f3f3f3f;
const double esp=1e-6;
const int maxn=1e6+5;
const int MOD=1e9+7;
const int mod=1e9+7;
int dir[5][2]={0,1,0,-1,1,0,-1,0};

ll inv[maxn*2],fac[maxn];
ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}
ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}
ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}
ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x);x=(x*x);}return res;}
void INV(){inv[1] = 1;for(int i = 2; i < maxn; i++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);}}
void Fac(){fac[0]=1;for(int i=1;i<=maxn;i++)fac[i]=(fac[i-1]*i)%MOD;}
ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}
ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}
ll inv2(ll b){return qpow(b,MOD-2);}
ll cal(ll m,ll n){if(m<n)return 0;return (fac[m]*inv[fac[n]]%MOD)%MOD*inv[fac[m-n]]%MOD;}
ll cals(ll m,ll n){if(m<n)return 0;return (fac[m]*inv1(fac[n])%MOD)%MOD*inv1(fac[m-n])%MOD;}

ll C[1005][1005];
void init()
{
    C[0][0]=1;

    for(int i=1;i<=50;i++)
        for(int j=0;j<=i;j++){
        	C[i][j]=(j==0)?1:(C[i-1][j]+C[i-1][j-1]);
		}
}

ll n,v1,v2,w;

int main()
{
    init();
    int t;
    //cout<<C[50][25]<<endl;
    cin>>t;
    while(t--)
    {
        scanf("%lld %lld %lld %lld",&n,&v1,&v2,&w);
        {
           ll temp1=n-v1-v2;//上
           ll temp2=n/2+1-v1; //下
           if(temp2<=0)temp2=0;
           ll sum=0;
           int flag=0;
           ll q= qpow(2,temp1);
           for(int i=temp2;i<=temp1&&!flag;i++)
           {
                sum+=C[temp1][i];
                if(sum*100>w*q)
                    flag=1;
           }
            if(flag)
               printf("GET A CRATE OF CHAMPAGNE FROM THE BASEMENT!\n");
            else if(v2>=ceil(n*1.0/2))
                printf("RECOUNT!\n");
            else
                printf("PATIENCE, EVERYONE!\n");
        }
    }
    return 0;
}

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转载于:https://www.cnblogs.com/sizaif/p/9078473.html